Free Flip-Book Mathematics class 12th by Study Innovations. 494 Pages

(4) Reflection (Image of a point) : Let (x, y) be any point, then its image with respect to

(i) x axis ⇒ (x,−y) (ii) y-axis ⇒ (−x, y) (iii) origin ⇒ (−x,−y) (iv) line y = x ⇒ (y, x)

Example: 21 The point (2,3) undergoes the following three transformation successively,
(i) Reflection about the line y = x .
Solution: (b)
Example: 22 (ii) Transformation through a distance 2 units along the positive direction of y-axis. [Roorkee 2000]
Solution: (a) (iii) Rotation through an angle of 45o about the origin in the anticlockwise direction.
Example: 23 The final coordinates of points are
Solution: (d)
(a)  1, 7  (b)  −1 , 7  (c)  1 , −7  (d) None of these
2 2 2 2 2 2

(i) The new position after reflection is (3,2)

(ii) After transformation, it is (3, 2+ 2), i.e, (3, 4)

(iii) Rotation makes it (3 cos 45o − 4 sin 45o,3 sin 45o + 4 cos 45o) , i.e.  −1 , 7 
2 2

Reflecting the point (2, –1) about y-axis, coordinate axes are rotated at 45o angle in negative direction without shifting the
origin. The new coordinates of the point are

(a)  −1 , −3  (b)  − 3 , 1  (c)  1, 3  (d) None of these
2 2 2 2 2 2

The new position after reflection is (–2, –1)

Rotation makes it [(−2) cos(−45o) + (−1) sin(−45o), − (−2) sin(−45o) + (−1) cos(−45o)] , i.e.,  −1 , −3
 2 
 2 

The point (3, 2) is reflected in the y-axis and then moved a distance 5 units towards the negative side of y-axis. The co-

ordinate of the point thus obtained are [DCE 1997]

(a) (3, –3) (b) (–3, 3) (c) (3, 3) (d) (–3, –3)

Reflection in the y-axis of the point (3,2) is (–3, 2) when it moves towards the negative side of y- axis through 5 units, then
the new position is (–3, 2– 5) =(– 3, – 3)

Y

Q 2 P
(–3, 2) (3, 2)
O
X′ X

3 Y′

R
(– 3, – 3)

Locus.
Locus : The curve described by a point which moves under given condition or conditions is called its locus.
Equation to the locus of a point : The equation to the locus of a point is the relation, which is satisfied by

the coordinates of every point on the locus of the point.
Algorithm to find the locus of a point

Step I : Assume the coordinates of the point say (h, k) whose locus is to be found.
Step II : Write the given condition in mathematical form involving h , k.
Step III : Eliminate the variable (s), if any.
Step IV : Replace h by x and k by y in the result obtained in step III. The equation so obtained is the locus of
the point which moves under some stated condition (s)

Note :  Locus of a point P which is equidistant from the two point A and B is a straight line and is a

perpendicular bisector of line AB.

 In above case if PA = kPB where k ≠ 1 , then the locus of P is a circle.
 Locus of P if A and B is fixed.

(a) Circle, if ∠APB = constant (b) Circle with diameter AB , if ∠APB = π
(c) Ellipse, if PA +PB = constant 2

(d) Hyperbola, if PA – PB = constant

Example: 24 Let A (2, – 3) and B( – 2, 1) be vertices of triangle ABC. If the centroid of this triangle moves on the line 2x + 3y = 1 , then
Solution: (d)
the locus of the vertex C is the line [AIEEE 2004]

(a) 3x − 2y = 3 (b) 2x − 3y = 7 (c) 3x + 2y = 5 (d) 2x + 3y = 9

Let third vertex C be (α, β )

∴ Centroid =  2 − 2 + α , −3 + 1 + β  , i.e.  α , β − 2 
 3 3   3 3 

According to question, 2 α  + 3 β − 2  =1 ⇒ 2α + 3β −6 = 3 ⇒ 2α + 3β = 9
 3   3 

Hence, locus of vertex C is 2x + 3y = 9 .

Example: 25 The ends of a rod of length l move on two mutually perpendicular lines. The locus of the point on the rod which divides it
Solution: (c)
in the ratio 1 : 2 is [IIT 1987; Rajasthan PET 1997]

(a) 36x2 + 9y2 = 4l2 (b) 36x2 + 9y2 = l2 (c) 9x2 + 36y2 = 4l2 (d) None of these

AP : PB = 1 : 2 , then h = 1×0 + 2×a = 2a or a = 3h , Similarly b = 3k Y
1+ 2 3 2 B(0, b)

Now we have OA2 + OB2 = AB2 ⇒  3h  2 + (3k)2 l2 2
 2  b P(h, k)
=
1
Hence locus of P (h, k) is given by 9x2 + 36y2 = 4l2
O aA X

(a,0)

Example: 26 If A and B are two fixed points and P is a variable point such that PA + PB = 4 , then the locus of P is a/an
Solution: (b)
[IIT 1989; UPSEAT 2001]

(a) Parabola (b) Ellipse (c) Hyperbola (d) None of these

We know that, PA + PB = constant. Then locus of P is an ellipse.

Rectangular-cartesian-co-ordinates-1

Introduction .

Co-ordinates of a point are the real variables associated in an order to a point to describe its location in some

space. Here the space is the two dimensional plane. The work of describing the Y Quadrant I
position of a point in a plane by an ordered pair of real numbers can be done in (+,+)
different ways. Quadrant II
(–,+)

The two lines XOX' and YOY' divide the plane in four quadrants. XOY, X' X
YOX', X' OY', Y'OX are respectively called the first, the second, the third and the Quadrant III O Quadrant IV
fourth quadrants. We assume the directions of OX, OY as positive while the (+,–)
directions of OX', OY' as negative. (–,–)

Y'

Quadrant x-coordinate y-coordinate point
First quadrant + + (+,+)
Second quadrant – + (–,+)
Third quadrant – – (–,–)
Fourth quadrant + – (+,–)

Cartesian Co-ordinates of a Point .
This is the most popular co-ordinate system.

Let us consider two intersecting lines XOX' and YOY', which are perpendicular to each other. Let P be any

point in the plane of lines. Draw the rectangle OLPM with its adjacent sides X′ Y
OL,OM along the lines XOX', YOY' respectively. The position of the point P can
be fixed in the plane provided the locations as well as the magnitudes of OL, OM M x P(x, y)
are known.
y
Axis of x : The line XOX' is called axis of x.
O LX

Axis of y : The line YOY' is called axis of y.

Co-ordinate axes : x axis and y axis together are called axis of Y′
co-ordinates or axis of reference.

Origin : The point ‘O’ is called the origin of co-ordinates or the origin.

Oblique axes : If both the axes are not perpendicular then they are called as oblique axes.

Let OL = x and OM = y which are respectively called the abscissa (or x-coordinate) and the ordinate (or y-
coordinate). The co-ordinate of P are (x, y).

Note :  Co-ordinates of the origin is (0, 0).

 The y co-ordinate of every point on x-axis is zero. Y P(r,θ)

 The x co-ordinate of every point on y-axis is zero. X r X

Polar Co-ordinates . θ
Let OX be any fixed line which is usually called the initial line and O be a fixed point O

on it. If distance of any point P from the O is 'r' and ∠XOP = θ , then (r, θ ) are called the
polar co-ordinates of a point P.

If (x, y) are the cartesian co-ordinates of a point P, then

x = r cosθ ; y = r sinθ and r = x 2 + y 2 Y'

θ = tan −1  y 
 x 

Distance Formula .

The distance between two points P(x1, y1) and Q(x 2 , y2 ) is given by

PQ = (PR)2 + (QR)2 = (x2 − x1)2 + (y2 − y1)2 YQ

Note :  The distance of a point M(x0 , y0 ) from origin O (0, 0) P R
X′ O X

OM = (x 2 + y 2 ) .
0 0

 If distance between two points is given then use ± sign. Y′

 When the line PQ is parallel to the y-axis, the abscissa of point P and Q will be equal i.e, x1 = x2 ;

∴ PQ =| y2 − y1 |

 When the segment PQ is parallel to the x-axis, the ordinate of the points P and Q will be equal i.e.,

y1 = y2 . Therefore PQ =| x2 − x1 |

(1) Distance between two points in polar co-ordinates : Let O be the pole and OX be the initial line.
Let P and Q be two given points whose polar co-ordinates are (r1,θ1) and (r2,θ 2 )respectively.

Then OP = r1, OQ = r2 P(r1,θ1)
∠POX = θ1 and ∠QOX = θ 2

then ∠POQ = (θ1 − θ2) Q(r2,θ2)
X
(OP)2 + (OQ)2 − (PQ)2 r1 (θ1-θ2)
2OP. OQ O
In ∆POQ, from cosine rule cos(θ 1 −θ2) = θ1 r2
θ2

∴ (PQ)2 = r12 + r22 − 2r1r2 cos(θ1 − θ2) M

∴ PQ = r12 + r22 − 2r1r2 cos(θ1 − θ2)

Note :  Always taking θ1 and θ2 in radians.

Example: 1 If the point (x, y) be equidistant from the points (a + b,b − a) and (a − b,a + b) , then [MP PET 1983, 94]

(a) ax + by = 0 (b) ax − by = 0 (c) bx + ay = 0 (d) bx − ay = 0

Solution: (d) Let points P(x,y) , A (a + b, b − a), B(a − b, a + b) .

According to Question, PA = PB , i.e., PA2 = PB2
⇒ (a + b − x)2 + (b − a − y)2 = (a − b − x)2 + (a + b − y)2

⇒(a + b)2 + x 2 − 2x(a + b) + (b − a)2 + y 2 − 2y(b − a) = (a − b)2 + x2 − 2x(a − b) + (a + b)2 + y2 − 2y(a + b)

⇒ 2x(a − b − a − b) = 2y(b − a − a − b) ⇒ − 4bx = − 4ay ⇒ bx − ay = 0

Example: 2 If cartesian co-ordinates of any point are ( 3,1) , then its polar co-ordinates is

(a) (2,π / 3) (b) ( 2,π / 6) (c) (2,π / 6) (d) None of these

Solution: (c) We know that x = r cos θ , y = r sinθ

∴ 3 = r cosθ , 1 = r sinθ

r= ( 3)2 + (1)2 = 2 , θ = tan −1  1  = π / 6
3

Polar co-ordinates = (2,π / 6) .

Geometrical Conditions.
(1) Properties of triangles
(i) In any triangle ABC, AB + BC > AC and | AB − BC|< AC .

(ii) The ∆ABC is equilateral ⇔ AB = BC = CA .

(iii) The ∆ABC is a right angled triangle ⇔ AB2 = AC2 + BC2 or AC2 = AB2 + BC2 or BC2 = AB2 + AC2 .
(iv) The ∆ABC is isosceles ⇔ AB = BC or BC = CA or AB = AC .

(2) Properties of quadrilaterals D C
(i) The quadrilateral ABCD is a parallelogram if and only if θθ

(a) AB = DC, AD = BC , or (b) the middle points of BD and AC are the same, A B

In a parallelogram diagonals AC and BD are not equal and θ ≠ π .
2

(ii) The quadrilateral ABCD is a rectangle if and only if

(a) AB = CD, AD = BC and AC2 = AB2 + BC2 or, (b) AB = CD, AD = BC, AC = BD or, (c) the middle points

of AC and BD are the same and AC=BD. (θ ≠ π / 2 ) DC

θ

AB

(iii) The quadrilateral ABCD is a rhombus (but not a square) if and only if (a) AB = BC = CD = DA and
AC ≠ BD or, (b) the middle points of AC and BD are the same and AB = AD but AC ≠ BD . (θ = π / 2)

D C
θ

(iv) The quadrilateral ABCD is a square if and only if AB

(a) AB = BC = CD = DA and AC = BD or (b) the middle points of AC and BD are the same and AC = BD,

(θ = π / 2) , AB = AD . D C

Note :  Diagonals of square, rhombus, rectangle and parallelogram always θ

bisect each other.

 Diagonals of rhombus and square bisect each other at right angle. A B
 Four given points are collinear, if area of quadrilateral is zero.

Example: 3 ABC is an isosceles triangle. If the co-ordinates of the base are B(1,3) and C (– 2,7) the co-ordinates of vertex A can be

[Orissa JEE 2002]

(a) (1, 6) (b)  − 1 , 5  (c)  5 , 6  (d) None of these
 2   6 

Solution: (c) Let the vertex of triangle be A(x, y) .

Then the vertex A(x, y) is equidistant from B and C because ABC is an isosceles triangle, therefore

(x − 1)2 + (y − 3)2 = (x + 2)2 + (y − 7)2 ⇒ 6x − 8y + 43 = 0

Thus, any point lying on this line can be the vertex A except the mid point  − 1 , 5 of BC. Hence vertex A is  5 , 6 
 2   6 

Example: 4 The extremities of diagonal of parallelogram are the points (3, – 4) and (– 6,5) if third vertex is (– 2,1), then fourth vertex is
Solution: (b)
Example: 5 [Rajasthan PET 1987]
Solution: (c)
(a) (1,0) (b) (–1,0) (c) (1,1) (d) None of these
Example: 6
Solution: (a) Let A(3,−4) and C(−6, 5) be the ends of diagonal of parallelogram ABCD. Let B(−2,1) and D be (x, y), then mid points of

diagonal AC and BD coincide. So, x − 2 = −6 + 3 and y +1 = 5 −4
2 2 2 2

x = −1, y = 0 . ∴ Coordinates of D are (–1, 0)

The vertices A and D of square ABCD lie on positive side of x and y-axis respectively. If the vertex C is the point (12, 17),

then the coordinate of vertex B are (b) (15, 3) Y
(a) (14, 16) (d) (17, 12)
(c) (17, 5) M 12 C
Let the co-ordinate of B be (h, k) a
5 θ
a

Draw BL and CM perpendicular to x-axis and y-axis. D

∴ a cosθ = CM = OD = AL = 12 12 θ a aB
and a sinθ = DM = OA = BL = 5 O 5A
∴ k = BL = DM = OM − OD = 17 − 12 = 5 5

θ X
12 L

∴ h = OL = OA + AL = 5 + 12 = 17

Hence, Point B is (17, 5). [AIEEE 2002]
A triangle with vertices (4, 0); (–1, –1); (3, 5) is

(a) Isosceles and right angled (b) Isosceles but not right angled

(c) Right angled but not isosceles (d) Neither right angled nor isosceles

Let A (4,0); B(−1,−1); C(3,5) then

AB = 26, AC = 26 , BC = 52 ; i.e. AB = AC
So triangle is isosceles and also (BC)2 = (AB)2 + (AC)2 . Hence ∆ ABC is right angled isosceles triangle.

Section Formulae.

If P(x, y) divides the join of A(x1, y1) and B(x 2 , y2 ) in the ratio m1 : m2 (m1,m2 > 0)

(1) Internal division : If P(x, y)divides the segment AB internally in the ratio of m1 : m2

⇒ PA = m1 (x2,y2)
PB m2 B

The co-ordinates of P(x,y) are P (x, y)

x = m1 x 2 + m2 x1 and y = m1 y 2 + m2y1 (x1,y1)
m1 + m2 m1 + m2 A

(2) External division : If P(x,y) divides the segment AB externally in the ratio of m1 : m2

⇒ PA = m1 P(x, y)
PB m2
(x2, y2)
m1 x 2 − m2 x1 m1 y 2 − m2y1
The co-ordinates of P(x,y) are x = m1 − m2 and y = m1 − m2 (x1,y1) A B

Note :  If P(x,y) divides the join of A(x1, y1 ) and B(x 2 , y2 ) in the ratio λ : 1(λ > 0) , then

x = λx2 ± x1 ; y = λy2 ± y1 . Positive sign is taken for internal division and negative sign is taken
λ ±1 λ ±1

for external division.

 The mid point of AB is  x1 + x2 , y1 + y2  [Here m1 : m2 :: 1 : 1 ]
 2 2 

 For finding ratio, use ratio λ : 1 . If λ is positive, then divides internally and if λ is negative, then

divides externally.

 Straight line ax + by + c = 0 divides the join of points A(x1, y1) and B(x2, y2) in the

ratio  − ax1 + by1 + c  .
ax 2 + by2 + c

If ratio is –ve then divides externally and if ratio is +ve then divides internally.

Example: 7 The co-ordinate of the point dividing internally the line joining the points (4, –2) and (8, 6) in the ratio 7: 5 will be

Solution: (c) [AMU 1979; MP PET 1984]
Example: 8
Solution: (b) (a) (16, 18) (b) (18, 16) (c)  19 , 8  (d)  8 , 19 
Example: 9  3 3   3 3 
Solution: (a)
Let point (x, y) divides the line internally.

Then x = m1 x 2 + m2 x1 = 7(8) + 5(4) = 19 , y = m1y2 + m2y1 = 7(6) + 5(−2) = 8 .
m1 + m2 12 3 m1 + m2 12 3

The line x + y = 4 divides the line joining the points (–1,1) and (5, 7) in the ratio [IIT 1965, UPSEAT 1999]

(a) 2 : 1 (b) 1 : 2 Internally (c) 1 : 2 Externally (d) None of these

Required ratio = −  ax1 + by1 + c  = –  −1 + 1 − 4  = 4 = 1 (Internally)
ax2 + by2 + c  5+7− 4  8 2

The line joining points (2, –3) and (–5, 6)is divided by y-axis in the ratio [MP PET 1999]

(a) 2 : 5 (b) 2 : 3 (c) 3 : 5 (d) 1 : 2

Let ratio be k :1 and coordinate of y-axis are (0, b). Therefore, 0 = k(−5) + 1(2) ⇒ k = 2
k +1 5

Some points of a Triangle.

(1) Centroid of a triangle : The centroid of a triangle is the point of A(x1,y1)
intersection of its medians. The centroid divides the medians in the ratio 2 : 1

(Vertex : base) 2 E
11 2
If A(x1, y1 ), B(x2, y2) and C(x 3 , y3 ) are the vertices of a triangle. If G be the F 2 G1
centroid upon one of the median (say) AD, then AG : GD = 2 : 1 B C
(x2,y2) D (x3,y3)
 x1 + x2 + x3 y1 + y2 + y3 
⇒ Co-ordinate of G are  3 , 3 

Example: 10 The centroid of a triangle is (2,7)and two of its vertices are (4, 8) and (–2, 6) the third vertex is [Kerala (Engg.) 2002]

(a) (0, 0) (b) (4, 7) (c) (7, 4) (d) (7, 7)

Solution: (b) Let the third vertex (x, y)

2= x + 4 − 2 , 7 = y + 8 + 6 , i.e. x = 4, y=7
3 3

Hence third vertex is (4, 7).

(2) Circumcentre : The circumcentre of a triangle is the point of intersection of the perpendicular bisectors of
the sides of a triangle. It is the centre of the circle which passes through the
vertices of the triangle and so its distance from the vertices of the triangle is the A(x1,y1)

same and this distance is known as the circum-radius of the triangle.

Let vertices A, B, C of the triangle ABC be (x1, y1),(x2, y2) and (x 3 , y3 )and FE

let circumcentre be O(x, y) and then (x, y) can be found by solving O

(OA)2 = (OB)2 = (OC)2 (x2,y2)B D C(x3,y3)

i.e., (x − x1)2 + (y − y1)2 = (x − x2)2 + (y − y2)2 = (x − x3)2 + (y − y3)2

Note :  If a triangle is right angle, then its circumcentre is the mid point of hypotenuse.

 If angles of triangle i.e., A, B, C and vertices of triangle A(x1, y1), B(x2, y2) and C (x 3 , y3 ) are given,
then circumcentre of the triangle ABC is

 x1 sin 2A + x 2 sin 2B +x 3 sin 2C , y1 sin 2A + y 2 sin 2B + y3 sin 2C 
sin 2A + sin 2B + sin 2C sin 2A + sin 2B + sin 2C

Example: 11 If the vertices of a triangle be (2, 1); (5, 2) and (3,4) then its circumcentre is [IIT 1964]
Solution: (b)
(a)  13 , 9  (b)  13 , 9  (c)  9 , 13  (d) None of these
 2 2   4 4   4 4 

Let circumcentre be O (x, y) and given points are A(2,1); B(5, 2); C(3, 4) and OA2 = OB2 = OC2

∴ (x − 2)2 + (y − 1)2 = (x − 5)2 + (y − 2)2 .....(i)

and (x − 2)2 + (y − 1)2 = (x − 3)2 + (y − 4)2 .....(ii)

On solving (i) and (ii), we get x = 13 , y = 9
4 4
A
(3) Incentre : The incentre of a triangle is the point of intersection of internal F IE
bisector of the angles. Also it is a centre of a circle touching all the sides of a triangle. B DC

Co-ordinates of incentre  ax1 + bx2 + cx3 , ay1 + by2 + cy3 
 a+b+c a+b+c 

Where a, b, c are the sides of triangle ABC.

(4) Excircle : A circle touches one side outside the triangle and other two extended sides then circle is known
as excircle. Let ABC be a triagle then there are three excircles with three excentres.

Let I1, I 2 , I 3 opposite to vertices A,B and C respectively. If vertices of triangle are I3 A I2
A(x1, y1), B(x2, y2) and C (x3, y3) then

I1 ≡  − ax 1 + bx 2 + cx 3 , − ay1 + by2 + cy3  BC
− a+b+c −a+b+c I1

I2 ≡  ax 1 − bx 2 + cx 3 , ay1 − by2 + cy 3  , I3 ≡  ax1 + bx2 − cx3 , ay1 + by2 − cy3 
a−b+c a−b+c  a+b−c a+b−c 

Note :  Angle bisector divides the opposite sides in the ratio of remaining sides e.g. BD = AB = c
DC AC b

 Incentre divides the angle bisectors in the ratio (b + c) : a,(c + a) : b and (a + b) : c

 Excentre : Point of intersection of one internal angle bisector and other two external angle bisector

is called as excentre. There are three excentres in a triangle. Co-ordinate of each can be obtained by

changing the sign of a,b,c respectively in the formula of in-centre.

Example: 12 The incentre of the triangle with vertices (1, 3),(0,0) and (2, 0) is [IIT Screening 2000]
Solution: (d)
(a) 1, 3  (b)  2 , 1  (c)  2 , 3  (d) 1, 1 
2  3 3  3 2  3

 Here AB = BC = CA (A1,

∴ The triangle is equilateral .

So, the incentre is the same as the centroid.

∴ Incentre =  1 + 0 + 2 , 3 + 0 + 0  = 1, 1  . 2 2
 3 3  3 C
(2
60o

B 2
(0,

(5) Orthocentre : It is the point of intersection of perpendiculars drawn from vertices on opposite sides (called
altitudes) of a triangle and can be obtained by solving the equation of any two
A (x1, y1)
altitudes.

Here O is the orthocentre since AE ⊥ BC , BF ⊥ AC and CD ⊥ AB D OF
then OE ⊥ BC, OF ⊥ AC, OD ⊥ AB

Solving any two we can get coordinate of O.

B C
(x2, y2) E (x3, y3)

Note :  If a triangle is right angled triangle, then orthocentre is the point where right angle is formed.

• If the triangle is equilateral then centroid, incentre, orthocentre, circum-centre coincides.
• Orthocentre, centroid and circum-centre are always collinear and centroid divides the line

joining orthocentre and circum-centre in the ratio 2 : 1
• In an isosceles triangle centroid, orthocentre, incentre, circum-centre lie on the same line.

Example: 13 The vertices of triangle are (0, 3) (– 3, 0) and (3, 0). The co-ordinate of its orthocentre are [AMU 1991; DCE 1994]
Solution: (c)
(a) (0, – 2) (b) (0, 2) (c) (0, 3) (d) (0, –3)

Here AB ⊥ BC .

In a right angled triangle, orthocentre is the point where right angle is formed. B (0,3)
∴ Orthocentre is (0, 3)

C O A
(– 3, 0) (3, 0)

Example: 14 If the centroid and circumcentre of triangle are (3, 3); (6, 2), then the orthocentre is [DCE 2000]
Solution: (d)
(a) (9, 5) (b) (3, –1) (c) (– 3, 1) (d) (– 3, 5)

Let orthocentre be (α, β ) . We know that centroid divides the line joining orthocentre and circumcentre in the ratio 2 : 1

∴3 = 2(6) + 1(α ) ⇒ α = −3 , 3 = 2(2) + 1(β ) ⇒ β = 5
2 + 1 2 + 1

Hence orthocentre is (–3, 5).

Area of some Geometrical figures.

(1) Area of a triangle : The area of a triangle ABC with vertices A(x1,y1); B(x2,y2) and C(x3, y3) . The area

of triangle ABC is denoted by ‘∆’and is given as A (x1, y1)

1 x1 y1 1 1
2 x2 y2 2
∆ = x3 y3 1 = (x1(y2 − y3) + x2(y3 − y1) + x3(y1 − y2)
1

In equilateral triangle B C
(x2, y2) (x3, y3)
(i) Having sides a, area is 3 a2 .
4

(ii) Having length of perpendicular as 'p' area is (p 2 ) .
3

Note :  If a triangle has polar co-ordinates (r1,θ1), (r2,θ2) and (r3,θ3) then its area

∆ = 1 [r1r2 sin(θ 2 − θ1) + r2r3 sin(θ3 − θ2) + r3r1 sin(θ1 − θ3)]
2

 If area is a rational number. Then the triangle cannot be equilateral.

(2) Collinear points : Three points A(x1, y1 ); B(x 2 , y2 ); C(x 3 , y3 ) are collinear. If area of triangle is zero,

i.e., (i) ∆=0 ⇒ 1 x1 y1 1 x1 y1 1
2 x2 y2 1 =0 ⇒ x2 y2 1 =0
x3 y3 1 x3 y3 1

(ii) AB + BC = AC or AC + BC = AB or AC + AB = BC

(3) Area of a quadrilateral : If (x1, y1);(x2, y2);(x3, y3) and (x 4 , y4 ) are vertices of a quadrilateral, then its

Area = 1 [(x 1 y 2 − x 2y1) + (x 2y3 − x3y2 ) + (x3y4 − x4 y3 ) + (x4 y1 − x1y4 )]
2

Note :  If two opposite vertex of rectangle are (x1, y1) and (x2, y2) , then its area is

(y2 − y1)(x2 − x1) .

 It two opposite vertex of a square are A(x1, y1) and C(x2, y2) , then its area is

= 1 AC2 = 1 [(x 2 − x1)2 + (y2 − y1 )2 ]
2 2

(4) Area of polygon : The area of polygon whose vertices are (x1, y1);(x 2 , y2 );(x 3 , y3 );....(xn,yn ) is

= 1 |{(x 1 y 2 − x 2y1) + (x 2y3 − x 3 y2 ) + .... + (xn y1 − x1yn )}|
2
Or Stair method : Repeat first co-ordinates one time in last for down arrow use positive sign and for up
arrow use negative sign.

x1 y1

x2 y2

∴ Area of polygon = 1 | x3 y3 |= 1 |{(x1y2 + x2y3 + .... + xny1) − (y1x2 + y2x3 + .... + yn x1 )}|
2 : : 2
Example: 15
Solution: (d) : :
Example: 16
xn yn
Solution: (a) x1 y1
Example: 17
Solution: (a) The area of the triangle formed by the points (a,b + c),(b,c + a),(c,a + b) is

Example: 18 [IIT 1963; EAMCET 1982; Rajasthan PET 2003]
Solution: (b,c)
(a) abc (b) a2 + b2 + c2 (c) ab + bc + ca (d) 0

a b+c 1a a+b+c 1a 1 1
1 c+a 1 b+c+a a+b+c 1 1 =0
Area = 2 b a+b 1 = 2 b c+a+b 1 , (Applying C2 → C1 + C2 ) = 2 b 1 1
c 1 c 1 c

Three points are A(6, 3), B(– 3, 5), C(4, – 2) and P (x, y) is a point, then the ratio of area of ∆PBC and ∆ABC is

[IIT 1983]

(a) x+y−2 (b) x−y+2 (c) x−y−2 (d) None of these
7 2 7

Area of ∆PBC 1 [− 3(−2 − y) + 4(y − 5) + x(5 + 2)] = 7x + 7y − 14 x+y−2
Area of ∆ABC 2 2) − 3(−2 − 3) + 4(3 − 5)] 49 7
= 1 =
2
[6(5 +

If the points (2K, K)(K, 2K) and (K, K) with K > 0 enclose triangle of area 18 square units then the centroid of triangle is
equal to

(a) (8, 8) (b) (4, 4) (c) (– 4, – 4) (d) (4 2, 4 2)

∆ = 1 2K K 1 K2 = 18 ⇒ K =± 6 . Consider K = +6 because K > 0, then the points (12, 6) (6,12) and (6,6).
2 K 2K 1 = 18 ⇒ 2
K K 1

Hence, centroid =  12 +6 +6,6 + 12 + 6  = (8, 8)
 3 3 

If the points (x+1, 2); (1,x + 2);  x 1 1 , x 2 1  are collinear, then x is [Rajasthan PET 2002]
 + + 

(a) 4 (b) 0 (c) – 4 (d) None of these

Let A = (x + 1, 2) ; B = (1, x + 2) ; C =  x 1 1 , x 2 1  . A, B, C are collinear, if area of ∆ABC = 0
 + + 

x +1 2 1 x −x 0

⇒ 1 x+2 1 =0⇒ 1 x+2 1 =0 (R1 → R1 − R2)
1 2 1 2
x +1 x +1 1 x +1 x +1 1

x 00

⇒ 1 x+3 1 =0 (C2 → C1 + C2) ⇒ x2(x + 4) = 0 ⇒ x = 0, − 4
1 3
x +1 x +1 1

Example: 19 The points (1, 1); (0, sec 2 θ ); (cosec 2θ ,0) are collinear for [Roorkee 1963]
Solution: (b)
(a) θ = nπ / 2 (b) θ ≠ nπ / 2 (c) θ = nπ (d) None of these

1 1 1 1
2 sec 2 θ 1 = 0 ⇒ 1(sec 2 θ ) + 1(cosec 2θ ) + 1(−cosec 2θ .sec 2 θ ) = 0
The given points are collinear, if Area of ∆ = 0
cosec 2θ 0 1

⇒ 1 1 1 =0 1 1 =0 ⇒0=0
cos2 θ + sin2 θ − sin2 θ . cos2 θ ⇒ sin2 θ .cos2 θ − sin2 θ .cos2 θ

Therefore the points are collinear for all value of θ , except only θ = nπ because at θ = nπ , sec2 θ =∞ (Not defined).
2 2

Example: 20 The points (0, 8/3) (1, 3) and (82, 30) are the vertices of [IIT 1983; Rajasthan PET 1988]
Solution: (d)
(a) An equilateral triangle (b) An isosceles triangle

(c) A right angled triangle (d) None of these

Here A = (0, 8 / 3), B = (1,3) and C = (82, 30)

AB = 1+1/9 = 10 / 9 , BC = (81)2 + (27)2 = 27 10 = 81 10 , AC = (82)2 + (30 − 8 / 3)2 = 82 10
9 9

Since AB + BC = (1 + 81) 10 = 82 10 = AC . ∴ Points A, B, C are collinear.
9 9

Transformation of Axes .

(1) Shifting of origin without rotation of axes : Let P ≡ (x, y) with respect to axes OX and OY.

Let O' ≡ (α, β ) with respect to axes OX and OY and let P ≡ (x' , y' ) with respect to axes O'X' and O'Y', where

OX and O'X' are parallel and OY and O'Y' are parallel. Y
Then x = x'+α, y = y' + β or x' = x − α, y' = y − β
Y′
Thus if origin is shifted to point (α, β ) without rotation of axes, then new
equation of curve can be obtained by putting x + α in place of x and y + β in (α,β ) P(x, y)
place of y. (x', y')
O′ y′
O
x′ X′

X

(2) Rotation of axes without changing the origin : Let O be the origin. Let P ≡ (x, y) with respect to axes

OX and OY and let P ≡ (x' , y' ) with respect to axes OX′ and OY′ where Y' Y
∠X' OX = ∠YOY' = θ

then x = x' cosθ − y' sinθ P (x,y)
y = x' sinθ + y' cosθ (yx′′,y′)

and x' = x cosθ + y sinθ θ y θ X'
y' = −x sinθ + y cosθ
x′

O xθ X

The above relation between (x, y) and (x' , y' ) can be easily obtained with the help of following table

x↓ y↓

x' → cos θ sinθ

y' → − sinθ cos θ

(3) Change of origin and rotation of axes : If origin is changed to Y
O'(α, β ) and axes are rotated about the new origin O' by an angle θ in the Y'

anticlock-wise sense such that the new co-ordinates of P(x, y) become θ

(x' , y' ) then the equations of transformation will be x = α + x' cosθ − y' sinθ O′ P(x, y)
O (x', y')
and y = β + x' sinθ + y' cosθ X'

θ

X

(4) Reflection (Image of a point) : Let (x, y) be any point, then its image with respect to

(i) x axis ⇒ (x,−y) (ii) y-axis ⇒ (−x, y) (iii) origin ⇒ (−x,−y) (iv) line y = x ⇒ (y, x)

Example: 21 The point (2,3) undergoes the following three transformation successively,
(i) Reflection about the line y = x .
Solution: (b)
Example: 22 (ii) Transformation through a distance 2 units along the positive direction of y-axis. [Roorkee 2000]
Solution: (a) (iii) Rotation through an angle of 45o about the origin in the anticlockwise direction.
Example: 23 The final coordinates of points are
Solution: (d)
(a)  1, 7  (b)  −1 , 7  (c)  1 , −7  (d) None of these
2 2 2 2 2 2

(i) The new position after reflection is (3,2)

(ii) After transformation, it is (3, 2+ 2), i.e, (3, 4)

(iii) Rotation makes it (3 cos 45o − 4 sin 45o,3 sin 45o + 4 cos 45o) , i.e.  −1 , 7 
2 2

Reflecting the point (2, –1) about y-axis, coordinate axes are rotated at 45o angle in negative direction without shifting the
origin. The new coordinates of the point are

(a)  −1 , −3  (b)  − 3 , 1  (c)  1, 3  (d) None of these
2 2 2 2 2 2

The new position after reflection is (–2, –1)

Rotation makes it [(−2) cos(−45o) + (−1) sin(−45o), − (−2) sin(−45o) + (−1) cos(−45o)] , i.e.,  −1 , −3
 2 
 2 

The point (3, 2) is reflected in the y-axis and then moved a distance 5 units towards the negative side of y-axis. The co-

ordinate of the point thus obtained are [DCE 1997]

(a) (3, –3) (b) (–3, 3) (c) (3, 3) (d) (–3, –3)

Reflection in the y-axis of the point (3,2) is (–3, 2) when it moves towards the negative side of y- axis through 5 units, then
the new position is (–3, 2– 5) =(– 3, – 3)

Y

Q 2 P
(–3, 2) (3, 2)
O
X′ X

3 Y′

R
(– 3, – 3)

Locus.
Locus : The curve described by a point which moves under given condition or conditions is called its locus.
Equation to the locus of a point : The equation to the locus of a point is the relation, which is satisfied by

the coordinates of every point on the locus of the point.
Algorithm to find the locus of a point

Step I : Assume the coordinates of the point say (h, k) whose locus is to be found.
Step II : Write the given condition in mathematical form involving h , k.
Step III : Eliminate the variable (s), if any.
Step IV : Replace h by x and k by y in the result obtained in step III. The equation so obtained is the locus of
the point which moves under some stated condition (s)

Note :  Locus of a point P which is equidistant from the two point A and B is a straight line and is a

perpendicular bisector of line AB.

 In above case if PA = kPB where k ≠ 1 , then the locus of P is a circle.
 Locus of P if A and B is fixed.

(a) Circle, if ∠APB = constant (b) Circle with diameter AB , if ∠APB = π
(c) Ellipse, if PA +PB = constant 2

(d) Hyperbola, if PA – PB = constant

Example: 24 Let A (2, – 3) and B( – 2, 1) be vertices of triangle ABC. If the centroid of this triangle moves on the line 2x + 3y = 1 , then
Solution: (d)
the locus of the vertex C is the line [AIEEE 2004]

(a) 3x − 2y = 3 (b) 2x − 3y = 7 (c) 3x + 2y = 5 (d) 2x + 3y = 9

Let third vertex C be (α, β )

∴ Centroid =  2 − 2 + α , −3 + 1 + β  , i.e.  α , β − 2 
 3 3   3 3 

According to question, 2 α  + 3 β − 2  =1 ⇒ 2α + 3β −6 = 3 ⇒ 2α + 3β = 9
 3   3 

Hence, locus of vertex C is 2x + 3y = 9 .

Example: 25 The ends of a rod of length l move on two mutually perpendicular lines. The locus of the point on the rod which divides it
Solution: (c)
in the ratio 1 : 2 is [IIT 1987; Rajasthan PET 1997]

(a) 36x2 + 9y2 = 4l2 (b) 36x2 + 9y2 = l2 (c) 9x2 + 36y2 = 4l2 (d) None of these

AP : PB = 1 : 2 , then h = 1×0 + 2×a = 2a or a = 3h , Similarly b = 3k Y
1+ 2 3 2 B(0, b)

Now we have OA2 + OB2 = AB2 ⇒  3h  2 + (3k)2 l2 2
 2  b P(h, k)
=
1
Hence locus of P (h, k) is given by 9x2 + 36y2 = 4l2
O aA X

(a,0)

Example: 26 If A and B are two fixed points and P is a variable point such that PA + PB = 4 , then the locus of P is a/an
Solution: (b)
[IIT 1989; UPSEAT 2001]

(a) Parabola (b) Ellipse (c) Hyperbola (d) None of these

We know that, PA + PB = constant. Then locus of P is an ellipse.

Trigonometrical-ratios-functions-and-identities

Definitions.

(1) Angle : The motion of any revolving line in a plane from its initial position (initial side) to the final position

(terminal side) is called angle. The end point O about which the line rotates is called the B
vertex of the angle.
Terminal side
(2) Measure of an angle : The measure of an angle is the amount of rotation from
the initial side to the terminal side. OA
Initial side

(3) Sense of an angle : The sense of an angle is determined by the direction of
rotation of the initial side into the terminal side. The sense of an angle is said to be positive or negative according as

the initial side rotates in anticlockwise or clockwise direction to get the terminal side.

B Oθ A
θ A B Negative angle
O Positive angle

(4) Right angle : If the revolving ray starting from its initial position to final position describes one quarter of a
circle. Then we say that the measure of the angle formed is a right angle.

(5) Quadrants : Let X' OX and YOY' be two lines at right angles in the plane of the paper. These lines divide
the plane of paper into four equal parts. Which are known as quadrants. The lines
X' OX and YOY' are known as x-axis and y-axis. These two lines taken together are Y

known as the co-ordinate axes. II quadrant I quadrant

(6) Angle in standard position : An angle is said to be in standard position if X′ O X
its vertex concides with the origin O and the initial side concides with OX i.e., the

positive direction of x-axis. III quadrant IV quadrant

(7) Angle in a quadrant : An angle is said to be in a particular quadrant if the Y′
terminal side of the angle in standard position lies in that quadrant.

(8) Quadrant angle : An angle is said to be a quadrant angle if the terminal side concides with one of the axes.

System of Measurement of Angles

There are three system for measuring angles

(1) Sexagesimal or English system : Here a right angle is divided into 90 equal parts known as degrees.
Each degree is divided into 60 equal parts called minutes and each minute is further divided into 60 equal parts

called seconds. Therefore, 1 right angle = 90 degree (= 90o)

1o = 60 minutes (= 60')

1' = 60 second (= 60' ' )

(2) Centesimal or French system : It is also known as French system, here a right angle is divided into 100
equal parts called grades and each grade is divided into 100 equal parts, called minutes and each minute is further

divided into 100 seconds. Therefore,

1 right angle = 100 grades (= 100g )

1 grade = 100 minutes (= 100' )

1 minute = 100 seconds (= 100' ' )
(3) Circular system : In this system the unit of measurement is radian. One radian, written as 1c , is the
measure of an angle subtended at the centre of a circle by an arc of length equal to the radius of the circle.

P
OA

Consider a circle of radius r having centre at O. Let A be a point on the circle. Now cut off an arc AP whose
length is equal to the radius r of the circle. Then by the definition the measure of ∠AOP is 1 radian (= 1c ) .

Relation between Three Systems of Measurement of an Angle.

Let D be the number of degrees, R be the number of radians and G be the number of grades in an angle θ.

Now, 90o = 1 right angle ⇒ 1o = 1 right angle
90

⇒ Do = D right angles ⇒ θ = D right angles ……..(i)
90 90

Again, π radians = 2 right angles ⇒ 1 radian = 2 right angles
π

⇒ R radians = 2R right angles ⇒ θ = 2R right angles ……..(ii)
π π

and 100 grades = 1 right angle ⇒ 1 grade = 1 right angle
100

⇒ G grades = G right angles ⇒ θ = G right angles ……..(iii)
100 100

From (i), (ii) and (iii) we get, D = G = 2R
90 100 π

This is the required relation between the three systems of measurement of an angle.

Note :  On radian = 180o ⇒ π radians = 180o ⇒ 1 radian = 57o 17′44.8′′ ≈ 57o17′45′′ .

π

Relation between an Arc and an Angle.

If s is the length of an arc of a circle of radius r, then the angle θ (in radians) subtended B
θC s
by this arc at the centre of the circle is given by θ = s or s = rθ i.e., arc = radius × angle Or A
r

in radians

Sectorial area : Let OAB be a sector having central angle θ C and radius r. Then area of the sector OAB is

given by 1 r 2θ .
2

Important Tips

 The angle between two consecutive digits in a clock is 30o (= π/6 radians). The hour hand rotates through an angle of 30o in one hour.
 The minute hand rotate through an angle of 6o in one minute.

Example: 1 The circular wire of radius 7 cm is cut and bend again into an arc of a circle of radius 12 cm. The angle subtended by an
Solution: (b)
Example: 2 arc at the centre of the circle is [Kerala (Engg.) 2002]
Solution: (d)
Example: 3 (a) 50o (b) 210o (c) 100o (d) 60o
Solution: (a)
Given the diameter of circular wire = 14 cm. Therefore length of wire = 14π cm
Example: 4
Solution: (b) Hence, required angle = Arc = 14π = 7π radian ⇒ 7π 180 o = 210o .
Example: 5 Radius 12 6 6 ×
Solution: (c)
π

The degree measure corresponding to the given radian  2π c
 15 

(a) 21o (b) 22o (c) 23o (d) 24o

We have, π radians = 180o

∴ 1c = 180 o ; ∴  2π c =  2π × 180 o = 24o .
 π   15   15 π 

The angles of a quadrilateral are in A.P. and the greatest angle is 120o, the angles in radians are

(a) π , 4π , 5π , 2π (b) π , π , 2π , 3π (c) 5π , 8π , 11π , 12π (d) None of these
3 9 9 3 3 2 3 3 18 18 18 18

Let the angles in degrees be α − 3δ ,α − δ ,α + δ ,α + 3δ

Sum of the angles = 4α = 360o ∴ α = 90o

Also greatest angle = α + 3δ = 120o , Hence, 3δ = 120o − α = 120o − 90o = 30o ∴ δ = 10o

Hence the angles are 90o − 30o,90o − 10o,90o + 10o and 90o + 30o

That is, the angles in degrees are 60o , 80o ,100o and 120o

∴ In terms of radians the angles are 60 × π , 80 × π , 100 × π and 120 × π that is π , 4π , 5π and 2π .
180 180 180 180 3 9 9 3

The minute hand of a clock is 10 cm long. How far does the tip of the hand move in 20 minutes

(a) 10π (b) 20π (c) 30π (d) 40π
3 3 3 3

We know that the tip of the minute hand makes one complete round in one hour i.e. 60 minutes since the length of the
hand is 10 cm. the distance moved by its tip in 60 minutes = 2π × 10cm = 20π cm

Hence the distance in 20 minutes = 20π × 20cm = 20π cm .
60 3

The angle subtended at the centre of radius 3 metres by the arc of length 1 metre is equal to [UPSEAT 1973]

(a) 20o (b) 60o (c) 1/3 radian (d) 3 radian

Required angle = Arc = 1 radian .
radius 3

Trigonometrical Ratios or Functions.

In the right angled triangle OMP, we have base = OM = x, perpendicular =PM = y and hypotenues = OP =r.
We define the following trigonometric ratio which are also known as trigonometric function.

sinθ = Perpendicular = y cosθ = Base = x Y
Hypotenues r Hypotenues r
A
tan θ = Perpendicular = y cot θ = Base = x , P(x, y)
Base x Perpendicular y

sec θ = Hypotenues = r cosecθ = Hypotenues = r ry
Base x Perpendicular y

(1) Relation between trigonometric ratio (function) θ MX
Ox
(i) sinθ .cosecθ = 1 (ii) tanθ .cotθ = 1

(iii) cosθ .secθ = 1 (iv) tanθ = sinθ (v) cotθ = cosθ
cosθ sinθ

(2) Fundamental trigonometric identities

(i) sin2 θ + cos2 θ = 1 (ii) 1 + tan2 θ = sec 2 θ (iii) 1 + cot2 θ = cosec 2θ

Important Tips

• If x = secθ + tanθ , then 1 = sec θ − tanθ .
x

• If x = coescθ + cotθ , then 1 = cosecθ − cotθ .
x

(3) Sign of trigonometrical ratios or functions : Their signs depends on the quadrant in which the terminal
side of the angle lies.

(i) In first quadrant : x > 0, y > 0 ⇒ sinθ = y > 0, cosθ = x > 0, tanθ = y > 0, cosecθ = r > 0,
r r x y

sec θ = r > 0 and cot θ = x > 0. Thus, in the first quadrant all trigonometric functions are positive.
x y

(ii) In second quadrant : x < 0, y > 0 ⇒ sinθ = y > 0, cosθ = x < 0, tanθ = y < 0, cosecθ = r > 0,
r r x y

sec θ = r <0 and cot θ = x < 0. Thus, in the second quadrant sin and cosec function are positive and all others
x y

are negative.

(iii) In third quadrant : Y

x < 0, y < 0 ⇒ sin θ = y < 0, cos θ = x < 0, tanθ = y > 0, cosecθ = r < 0 II quadrant I quadrant
r r x y S A

, sec θ = r < 0 and cot θ = x > 0. Thus, in the third quadrant all x < 0, y > 0 x > 0, y > 0
x y sin and cosec are All are positive

X’ positive IV quadrant X
III quadrant O C
trigonometric functions are negative except tangent and cotangent.
T

(iv) In fourth quadrant : x > 0, y < 0 ⇒ sinθ = y < 0, x < 0, y < 0 x > 0, y < 0
r tan and cot cos and sec
are positive are positive

Y’

cosθ = x > 0, tan θ = y < 0, cosecθ = r <0 , sec θ = r >0 and cot θ = x <0 Thus, in the fourth quadrant all
r x y x y

trigonometric functions are negative except cos and sec.

In brief : A crude aid to memorise the signs of trigonometrical ratio in different quadrant. "Add Sugar To

Coffee".

Important Tips

o First determine the sign of the trigonometric function.
o If θ is measured from X′OX i.e., {(π ± θ, 2π – θ)} then retain the original name of the function.

o If θ is measured from Y ′OY i.e., π ± θ, 3π ± θ  , then change sine to cosine, cosine to sine, tangent to cotangent, cot to tan, sec
 2 
 2 

to cosec and cosec to sec.

(4) Variations in values of trigonometric functions in different quadrants : Let X' OX and YOY' be the
coordinate axes. Draw a circle with centre at origin O and radius unity.

Let M(x, y) be a point on the circle such that ∠AOM = θ then x = cosθ and y = sin θ ; − 1 ≤ cosθ ≤ 1 and
− 1 ≤ sinθ ≤ 1 for all values of θ .

II-Quadrant (S) I-Quadrant (A) Y
sinθ → decreases from 1 to 0 sinθ → increases from 0 to 1
cosθ → decreases from 0 to – 1 cosθ → decreases from 1 to 0 B (0,1)
tanθ → increases from – ∞ to 0 tanθ → increases from 0 to ∞ M (x, y)
cotθ → decreases from 0 to – ∞ cotθ → decreases from ∞ to 0
secθ → increases from – ∞ to – 1 secθ → increases from 1 to ∞ θ Xy
cosecθ → increases from 1 to ∞ cosecθ → decreases from ∞ to 1
X′ O X
III-Quadrant (T) IV-Quadrant (C) (–1, 0) A′ x N A (1, 0)
sinθ → decreases from 0 to – 1 sinθ → increases from – 1 to 0
cosθ → increases from – 1 to 0 cosθ → increases from 0 to 1 B′(0, –1)
tanθ → increases from 0 to ∞ tanθ → increases from – ∞ to 0 Y′
cotθ → decreases from ∞ to 0 cotθ → decreases from 0 to – ∞
secθ → decreases from – 1 to – ∞ secθ → decreases from ∞ to 1
cosecθ → increases from – ∞ to – 1 cosecθ → decreases from – 1 to – ∞

Note :  + ∞ and – ∞ are two symbols. These are not real number. When we say that tanθ increases from 0 to

∞ for as θ varies from 0 to π it means that tanθ increases in the interval  0, π  and it attains large positive values
2  2 

as θ tends to π . Similarly for other trigonometric functions.
2

Example: 6 If sinθ + cosecθ = 2 , then sin2 θ + cosec2θ = [UPSEAT 2002; MP PET 1992; MNR 1990]
Solution: (c)
(a) 1 (b) 4 (c) 2 (d) None of these

(sin2 θ + cosec2θ ) = (sinθ + cosecθ )2 − 2 sinθ .cosecθ = 22 − 2 = 2 .

Example: 7 If sinθ + cosθ = m and sec θ + cosecθ = n , then n(m + 1)(m − 1) equal to [MP PET 1986]
Solution: (c)
Example: 8 (a) m (b) n (c) 2m (d) 2n
Solution: (b)
n(m2 − 1) = (sec θ + cosecθ ).2 sinθ . cosθ [m2 = 1 + 2 sinθ . cosθ ]
Example: 9
Solution: (a) = sinθ + cosθ .2 sinθ . cosθ = 2m .
Example: 10 sinθ . cosθ
Solution: (c)
Example: 11 If tan θ = 1 x sin φ and tan φ = y sinθ , then x/y equal to [MP PET 1991]
Solution: (d) − x cos φ 1 − y cosθ
Example: 12
(a) sin φ (b) sinθ (c) sinφ (d) sin θ
Solution: (d) sinθ sinφ 1 − cosθ 1 − cos φ

Example: 13 x sin φ = tanθ − x cos φ. tanθ

x = sin φ tan θ = sin θ sin θ sin φ
+ cos φ tanθ cos φ + cosθ

Similarly, y = sin θ sin φ sin φ ; ∴ x = sinθ .
cos φ + cosθ y sin φ

The equation sec2 θ = 4 xy is only possible when [MP PET 1986; IIT 1996; Karnataka CET 1997; AMU 1987, 1991]
(x + y)2

(a) x = y (b) x < y (c) x > y (d) None of these

 cos 2 θ ≤ 1 ⇒ sec2 θ = 4 xy ≥1⇒ 4 xy ≥ (x + y)2 ⇒ (x − y)2 ≤ 0
(x + y)2

Which is possible only when x = y (x, y ∈ R)

1 − sinθ equals
1 + sinθ

(a) 0 (b) 1 (c) sec θ − tanθ (d) sec θ . tanθ

(1 − sinθ )2 = 1 − sin θ = sec θ − tanθ .
(1 − sin 2 θ ) cos θ

If tan A + cot A = 4, then tan 4 A + cot 4 A is equal to [Kerala (Engg.) 2002]
[UPSEAT 2003]
(a) 110 (b) 191 (c) 80 (d) 194

tan A + cot A = 4 ⇒ tan 2 A + cot 2 A + 2 tan A cot A = 16

⇒ tan2 A + cot2 A = 14 ⇒ tan 4 A + cot 4 A + 2 = 196 ⇒ tan 4 A + cot 4 A = 194 .

If sin x + cos x = 1 , then tan 2x is
5

(a) 25 (b) 7 (c) 25 (d) 24
17 25 7 7

sin x + cos x = 1 ⇒ sin 2 x + cos 2 x + 2 sin x cos x = 1
5 25

sin 2x = − 24 ⇒ cos 2x = − 7 ⇒ tan 2x = 24 .
25 25 7

If sin x = − 24 , then the value of tan x is [UPSEAT 2003]
25

(a) 24 (b) −24 (c) 25 (d) None of these
25 7 24

Solution: (b) cos x = 1 − sin2 x = 1 −  − 24 2 = 7 ⇒ tan x = sin x = −24 .
Example: 14  25  25 cos x 7
Solution: (b)
If tanθ + sec θ = ex, then cosθ equals [AMU 2002]
Example: 15
Solution: (d) (a) (e x + e−x ) (b) 2 (c) (e x − e−x ) (d) (e x − e−x )
2 (e x + e−x ) 2 (e x + e−x )
Example: 16
Solution: (d) tanθ + sec θ = ex ........(i)
Example: 17
Solution: (c) ∴ sec θ − tanθ = e−x ........(ii)
Example: 18
Solution: (d) From (i) and (ii), ⇒ 2 sec θ = ex + e−x ⇒ cos θ = ex 2 .
+ e−x

π ∞ ∞ ∞
2
x = cos2nφ, y = sin2nφ , z = cos2nφ sin2n φ ,then

n=0 n=0 n=0
∑ ∑ ∑For0 < φ < , if [IIT 1993]

(a) xyz = xz + y (b) xyz = xy + z (c) xyz = x + y + z (d) Both (b) and (c)

From s∞ = a
1−r

We get, x = 1 = 1 , y = 1 = 1 , z = 1 = 1 = xy
1 − cos2 φ sin2 φ sin2 cos2 φ cos2 φ xy − 1
1 − φ 1 − sin2 φ 1 − 1
xy

⇒ xyz − z = xy ⇒ xyz = xy + z ......(i)

Also, 1 + 1 = cos2 φ + sin2 φ =1 ⇒ x + y = xy ; From (i), xyz = x + y + z.
x y

If P = 1 + 2 sinθ and Q = 1 cos θ θ , then [MP PET 2001]
sinθ + cosθ + sin

(a) PQ = 1 (b) Q = 1 (c) Q − P = 1 (d) Q + P = 1
P

P +Q= 1+ 2 sinθ + cos θ
sinθ + cosθ 1 + sinθ

After solving, P + Q = 1 .

The value of 6(sin 6 θ + cos 6 θ ) − 9(sin 4 θ + cos 4 θ ) + 4 equals to [MP PET 2001, 1997]

(a) – 3 (b) 0 (c) 1 (d) 3

= 6(sin6 θ + cos6 θ ) − 9(sin4 θ + cos4 θ ) + 4

= 6[(sin2 θ + cos2 θ )3 − 3 sin2 θ cos2 θ (sin2 θ + cos2 θ )] − 9[(sin2 θ + cos2 θ )2 − 2 sin2 θ . cos2 θ ] + 4

= 6[1 − 3 sin2 θ cos2 θ ] − 9[1 − 2 sin2 θ cos2 θ ] + 4 = 6 − 9 + 4 = 1 .

sin θ + 1 cos θ equals to [Karnataka CET 1998]
1 − cotθ − tanθ
(d) cosθ + sinθ
(a) 0 (b) 1 (c) cosθ − sinθ

sin θ . sinθ ) + (1 cosθ . cosθ = sin2 θ + cos2 θ = cos2 θ − sin2 θ = cosθ + sinθ .
sinθ (1 − cotθ − tanθ ) cosθ (sinθ − cosθ ) (cosθ − sinθ ) (cos θ − sinθ )

Trigonometrical Ratios of Allied Angles.

Two angles are said to be allied when their sum or difference is either zero or a multiple of 90o .

(1) Trigonometric ratios of (–θ): Let a revolving ray starting from its Y

initial position OX, trace out an angle ∠XOA = θ . Let P(x, y) be a point on OA A
r P(x, y)
such that OP = r. Draw PM ⊥ from P on x-axis. Angle ∠XOA' = −θ in the
θ
clockwise sense. Let P' be a point on OA' such that OP' = OP. Clearly M and O –θ M

M′ coincide and ∆OMP is congruent to ∆OMP' then P' are (x, – y). r X
P′ (x, –y)
sin(−θ ) = −y ⇒ −y = − sin θ ; cos(−θ ) = x = cos θ ; tan(−θ ) = −y = − tan θ
r r r x

Taking the reciprocal of these trigonometric ratios; A′

cosec(−θ ) = − cosecθ , sec(−θ ) = sec θ and cot(−θ ) = − cotθ

Note :  A function f(x) is said to be an even function if f(−x) = f(x) for all x in its domain.

 A function f(x) is said to be an odd function if f(−x) = − f(x) for all x in its domain.

 sinθ , tanθ , cotθ , cosecθ are odd functions and cos θ , sec θ are even functions.

(2) Trigonometric function of (90 – θ ) : Let the revolving line, starting from P
OA, trace out any acute angle AOP, equal to θ. From any point P, draw PM ⊥ to OA. 90o–θ

Three angles of a triangle are together equal to two right angles, and since OMP is a

right angle, the sum of the two angles MOP and OPM is right angle. ∠OPM = 90o −θ 90o A
[When the angle OPM is consider, the line PM is the ‘base’ and MO is the O θ M

‘perpendicular’]

sin(90o −θ) = sin MPO = MO = cos AOP = cos θ , cos(90o −θ) = cos MPO = PM = sin AOP = sin θ
PO PO

tan(90o −θ) = tan MPO = MO = cot AOP = cot θ , cot(90o −θ) = cot MPO = PM = tan AOP = tan θ
PM MO

cosec(90 o − θ ) = cosec MPO = PO = sec AOP = sec θ , sec(90o −θ) = sec MPO = PO = cosec AOP = cosecθ
MO PM

(3) Trigonometric function of (90+θ ) : Let a revolving ray OA starting from its initial poisiton OX, trace out
an angle ∠XOA = θ and let another revolving ray OA′ starting from
the same initial position OX, first trace out an angle θ. So as to Y A
coincide with OA and then it revolves through an angle of 90o in A′

anticlockwise direction to form an angle ∠XOA' = 90o + θ . X′ (–y, x) P′ P(x, y) X
Let P and P' be points on OA and OA' respectively such that θθ
θ
OP = OP' = r . M′ O M

Draw perpendicular PM and PM' from P and P' respectively Y′
on OX . Let the coordinates of P be (x, y). Then OM = x and

PM = y clearly, OM' = PM = y and P' M' = OM = x .

So the coordinates of P' are –y, x

sin(90 + θ ) = M' P' = x = cosθ , cos(90 +θ) = OM' = −y = − sinθ
OP' r OP' r

tan(90 + θ ) = M' P' = x = −x = − cotθ , cot(90 +θ) = − tanθ , sec(90 + θ ) = −cosecθ , cosec(90 + θ ) = secθ
OM' −y y

Allied angles (−θ ) (90 – θ) (90 + θ ) (180 − θ ) (180 + θ ) (270 − θ ) (270 + θ) (360 − θ )
or or or
Trigo. Ratio or or or or
(π −θ) (π +θ) (2π −θ )
 π  π  3π − θ   3π 
 2 − θ   2 + θ   2   2 + θ 
 

sinθ – sinθ cosθ cosθ sinθ – sinθ – cosθ – cos θ – sinθ
– cosθ cosθ
cosθ cosθ sinθ – sinθ – cosθ tanθ – sinθ sinθ – tanθ
tan θ – tanθ cotθ – cotθ – tanθ cotθ – cotθ
π/2 2π
Important Tips 1 0
0 1
• sin nπ = 0, cos nπ = (−1)n ∞ 0

• sin(nπ + θ ) = (−1)n sinθ , cos( nπ + θ ) = (−1)n cosθ

• sin nπ n−1 , if n is odd
 2
+ θ  = (−1) 2 cosθ


= (−1)n / 2 sinθ , if n is even

 cos nπ + θ  n+1 if n is odd
 2 
= (−1) 2 sinθ ,

= (–1)n / 2 cosθ , if n is even

Trigonometrical Ratios for Various Angles.

θ 0 π/6 π/4 π/3 π 3π/2
0 –1
sinθ 0 1/2 1 / 2 3/2
–1 0
cosθ 1 3 / 2 1/ 2 1/2
0∞
tanθ 0 1/ 3 1 3

Trigonometrical Ratios in terms of Each other

sinθ cosθ tanθ cotθ secθ cosecθ
sinθ 1 − cos2 θ tan θ 1 sec2 θ − 1 1

sinθ 1 + tan2 θ 1 + cot2 θ sec θ cosecθ
1
cosθ 1 − sin2 θ cosθ 1 cot θ cosec2θ − 1
tanθ 1 + tan2 θ 1 + cot2 θ sec θ cosecθ
cot θ sin θ 1 − cos2 θ 1
1 − sin2 θ cos θ tanθ 1 sec2 θ − 1
1 − sin2 θ cos θ cot θ cosec2θ − 1
1 1 cosec2θ − 1
sin θ 1 − cos2 θ tan θ cotθ sec2 θ − 1
cosecθ
secθ 1 1 1 + tan2 θ 1 + cot2 θ secθ cosec2θ − 1
cosecθ cos θ cot θ
1 − sin2 θ 1 + tan2 θ sec θ cosecθ
1 1 tan θ 1 + cot2 θ sec2 θ − 1
1 − cos2 θ
sin θ

Important Tips

 Values for some standard angles

sin 15o = cos 75o = 3 −1; cos 15o = sin 75o = 3 +1 ; tan 15o = cot 75o = 2 − 3 ;

22 22

sin 18o = cos 72o = 5− 1 ; cos 36o = sin 54o = 5+ 1 ; tan 75o = cot 15o = 2 + 3
4 4

sin 22 1o = cos 67 1o = 2− 2, cos 22 1o = sin 67 1o = 2+ 2; cot 22 1o = tan 67 1o = 2 +1
2 2 2 2 2 2 2 2

tan 22 1o = cot 67 1o = 2 −1
2 2

Example: 19 sin 75o = [MNR 1979]

Solution: (b) (a) 2− 3 (b) 3 + 1 (c) − 3 − 1 (d) 3 − 1
Example: 20 2 22 22 22

Solution: (c) sin 75o = sin(45o + 30o) = sin 45o cos 30o + sin 30o cos 45o = 1 31 1 3 +1 .
Example: 21 × 2 + 2× = 22
Solution: (a)
Example: 22 2 2

Solution: (a) The value of cos A − sin A , when A = 5π is [MP PET 1990]
Example: 23 4
Solution: (b)
(a) 2 (b) 1 (c) 0 (d) 1
2

cos 5π − sin 5π ⇒ − cos π + sin π ⇒ − 1+ 1 =0.
4 4 4 4 2 2

tan A + cot(180o + A) + cot(90o + A) + cot(360o − A) equal to [MP PET 1992]

(a) 0 (b) 2 tan A (c) 2 cot A (d) 2(tan A − cot A)

tan A + cot A + (− tan A) + (− cot A) = 0 .

The value of cos 15o − sin 15o equal to [UPSEAT 1975; MP PET 1994; MP PET 2002]

(a) 1 (b) 1 (c) −1 (d) Zero
2 2 2

3 +1 − 3 −1 = 1.
2 2 22 2

3sin4  3π − α  + sin4 (3π + α ) − 2sin6  π + α  + sin6 (5π − α ) [IIT 1986]
  2   2
    (d) sin 4α + sin 6α

(a) 0 (b) 1 (c) 3

= 3[(− cosα)4 + (− sinα)4 ] − 2[ cos6 α + sin6 α ]

= 3[(cos 2 α + sin 2 α)2 − 2 sin 2 α cos 2 α] − 2[(cos 2 α + sin 2 α)3 − 3 cos 2 α sin 2 α (cos 2 α + sin 2 α)]

= 3 − 6 sin 2 α cos 2 α − 2 + 6 sin 2 α cos 2 α = 1.

Trick : Put α = 0, π ; then the value of expression remains constant i.e., it is independent of α.
2

Example: 24 Which of the following number is rational [IIT 1998]
Solution: (c)
(a) sin 15o (b) cos 15o (c) sin 15o. cos 15o (d) sin 15o. cos 75o
Example: 25
Solution: (c) sin 15o = sin(45o − 30o ) = 3 − 1 = irrational ⇒ cos 15o = cos(45o − 30o) = 3 + 1 = irrational
22 22
Example: 26
Solution: (b) sin 15o. cos 15o = 1 (2 sin 15 o cos 15o ) = 1 sin 30o = 1 = rational
2 2 4

sin 15o. cos 75o = sin 15o. sin 15o = sin 2 15 o =  3 − 1 2 = 4 −2 3 = irrational.
 2 2  8

If sin x + sin 2 x = 1 , then the value of cos12 x + 3 cos10 x + 3 cos 8 x + cos 6 x − 2 is [MP PET 2001]

(a) 0 (b) 1 (c) − 1 (d) 2

Since sin x + sin 2 x = 1 ⇒ sin x = 1 − sin 2 x = cos 2 x ........(i)

From given expression, cos 6 x(cos 6 x + 3 cos 4 x + 3 cos 2 x + 1) – 2 = cos 6 x(cos 2 x + 1)3 − 2

From (i) sin x = cos 2 x
∴ sin 3 x(sin x + 1)3 − 2 = (sin 2 x + sin x)3 − 2 = 1 − 2 = −1 .

If 4 sinθ = 3 cosθ then sec 2 θ equals to
4[1 − tan 2 θ ]

(a) 25 (b) 25 (c) 1 (d) 1
16 28 4

Given 4 sinθ = 3 cosθ ⇒ tan θ = 3
4

The given expression is sec2 θ 1 + tan2 θ = 1 + 9 25 .
4[1 − tan2 θ ] 4(1 − tan2 θ ) 16 28
= 9 =
41 16 
 − 

Formulae for the Trigonometric Ratios of Sum and Differences of Two Angles.

(1) sin(A + B) = sin A cos B + cos A sin B (2) sin(A − B) = sin A cos B − cos A sin B

(3) cos(A + B) = cos A cos B − sin A sin B (4) cos(A − B) = cos A cos B + sin A sin B

(5) tan(A + B) = tan A + tan B (6) tan(A − B) = tan A − tan B
1 − tan A tan B 1 + tan A tan B

(7) cot(A + B) = cot A cot B − 1 (8) cot(A − B) = cot A cot B + 1
cot A + cot B cot B − cot A

(9) sin(A + B). sin(A − B) = sin2 A − sin2 B = cos2 B − cos2 A

(10) cos(A + B).cos(A − B) = cos2 A − sin2 B = cos2 B − sin2 A

(11) tan A± tan B = sin A ± sin B = sin A cos B ± cos A sin B = sin(A ± B)  A ≠ nπ + π , B ≠ mπ 
cos A cos B cos A cos B cos A. cos B  2 

(12) cot A ± cot B = sin(B ± A)  A ≠ nπ , B ≠ mπ + π 
sin A. sin B  2 

Formulae for the Trigonometric Ratios of Sum and Differences of Three Angles.
(1) sin(A + B + C) = sin A cos B cos C + cos A sin B cos C + cos A cos B sin C − sin A sin B sin C

or sin (A + B + C) = cos A cos B cos C(tan A + tan B + tan C − tan A. tan B. tan C)

(2) cos(A + B + C) = cos A cos B cos C − sin A sin B cos C − sin A cos B sin C − cos A sin B sin C

cos(A + B + C) = cos A cos B cos C(1 − tan A tan B − tan B tan C − tan C tan A)

(3) tan(A + B + C) = tan A + tan B + tan C − tan A tan B tan C
1 − tan A tan B − tan B tan C − tan C tan A

(4) cot(A + B + C) = cot A cot B cot C − cot A − cot B − cot C
cot A cot B + cot B cot C + cotC.cot A − 1

In general;

(5) sin(A1 + A2 + ...... + An ) = cos A1 cos A2 ..... cos An (S1 − S3 + S5 − S7 + ...)
(6) cos(A1 + A2 + .... + An ) = cos A1 cos A2 ... cos An (1 − S2 + S4 − S6 ....)

(7) tan(A1 + A2 + ..... + An ) = S1 − S3 + S5 − S7 + ....
1 − S2 + S4 − S6 + ....

Where; S1 = tan A1 + tan A2 + .... + tan An = The sum of the tangents of the separate angles.

S2 = tan A1 tan A2 + tan A1 tan A3 + .... = The sum of the tangents taken two at a time.

S3 = tan A1 tan A2 tan A3 + tan A2 tan A3 tan A4 + ... = Sum of tangents three at a time, and so on.

If A1 = A2 = .... = An = A, then S1 = n tan A , S2 =nC2 tan 2 A , S3 =nC3 tan 3 A,....

(8) sin nA = cosn A(nC1 tan A −nC3 tan3 A +nC5 tan5 A − ....)

(9) cos nA = cos n A(1 −nC2 tan 2 A + nC4 tan 4 A − ...)

(10) tan nA = 1 nC1 tan A −nC3 tan3 A +nC5 tan5 A − ....
−nC2 tan 2 A +nC4 tan4 A −nC6 tan6 A + ...

(11) sin nA + cos nA = cosn A(1 +nC1 tan A −nC2 tan2 A −nC3 tan3 A +nC4 tan4 A +nC5 tan5 A −nC6 tan6 A − .....)

(12) sin nA − cos nA = cos n A(−1 + nC1 tan A + nC2 tan 2 A −nC3 tan 3 A −nC4 tan 4 A + nC5 tan 5 A + nC6 tan 6 A...)

(13) sin(α) + sin(α + β ) + sin(α + 2β ) + ..... + sin(α + (n − 1)β ) = sin{α + (n − 1)(β / 2)}. sin(nβ / 2)
sin(β / 2)

cosα + (n − 1) β . sinn β 
  2   2 
(14) cos(α) + cos(α + β ) + cos(α + 2β ) + .... + cos(α + (n − 1)β ) =
sin β 
 2 

Formulae to Transform the Product into Sum or Difference.

(1) 2 sin A cos B = sin(A + B) + sin(A − B) (2) 2 cos A sin B = sin(A + B) − sin(A − B)

(3) 2 cos A cos B = cos(A + B) + cos(A − B) (4) 2 sin A sin B = cos(A − B) − cos(A + B)

Let A + B = C and A − B = D

Then, A= C+D and B= C−D
2 2

Therefore, we find out the formulae to transform the sum or difference into product.

(5) sin C + sin D = 2 sin C + D cos C − D (6) sin C − sin D = 2 cos C + D sin C − D
2 2 2 2

(7) cos C + cos D = 2 cos C + D cos C − D (8) cos C − cos D = 2 sin C + D sin D − C = −2 sin C + D sin C − D
2 2 2 2 2 2

Important Tips

• sin(60 o − θ ). sinθ sin(60 o +θ) = 1 sin 3θ  cos(60 − θ ). cos θ cos(60 o +θ) = 1 cos 3θ
4 4

• tan(60o −θ ). tanθ tan(60o +θ )= tan 3θ

• cos A.cos 2A.cos 22 A.cos 23 A....... cos 2n−1 A = sin 2n A , if A = nπ
2n sin A

= 1, if A = 2nπ

= 1, if A = (2n + 1)π

Example: 27 cos12o − sin12o sin 147 o = [MP PET 1991]
cos12o + sin12o + cos147o
Solution: (d)
Example: 28 (a) 2 tan 33o (b) 1 (c) – 1 (d) 0
Solution: (d)
Example: 29 = 1 − tan 12o + tan 147o = tan(45 o − 12) + tan(180o − 33o ) = tan 33o + (− tan 33o ) = 0 .
1 + tan 12o

If sinθ1 + sinθ 2 + sinθ 3 = 3 , then cosθ1 + cosθ 2 + cosθ 3 = [EAMCET 1994]

(a) 3 (b) 2 (c) 1 (d) 0

We know | sinθ |≤ 1 ; So, each θ1,θ2 and θ 3 must be equal to π / 2

∴ cosθ1 + cosθ 2 + cosθ 3 = 0 .

cos A + cos(240o + A) + cos(240o − A) = [MP PET 1991]

(a) cos A (b) 0 (c) 3 sin A (d) 3 cos A

Solution: (b) cos A + [2 cos 240o cos A] = cos A + 2(− cos 60o ) cos A
Example: 30
Solution: (b) = cos A1 − 2 1  =0.
Example: 31   2 
Solution: (c)
Example: 32 sin sin 2 A − sin 2 B B = [MP PET 1993]
Solution: (b) A cos A − sin B cos
(d) cot(A + B)
Example: 33 (a) tan(A − B) (b) tan(A + B) (c) cot(A − B)
Solution: (d)
2(sin 2 A − sin 2 B) = 2 sin(A + B). sin(A − B) = 2 sin(A + B) sin(A − B) = tan(A + B) .
2 sin A cos A − 2 sin B cos B sin 2A − sin 2B 2 sin(A − B) cos(A + B)

The expression cos 2 (A − B) + cos 2 B − 2 cos(A − B) cos A cos B is

(a) Dependent on B (b) Dependent on A and B

(c) Dependent on A (d) Independent of A and B

cos 2 (A − B) + cos 2 B − cos(A − B)[cos(A − B) + cos(A + B)]

⇒ cos 2 B − cos(A − B) cos(A + B) ⇒ cos 2 B − (cos 2 A − sin 2 B) = 1 − cos 2 A

Trick : Put A = 90o and 0o the value is sin2 B + cos2 B = 1 and 0 again put B = 0o , 90o and the value is sin2 A and
sin 2 A means expression depends on A.

If tan α = m 1 and tan β 1 then α+β = [IIT 1978]
m+ = 2m + 1

(a) π (b) π (c) π (d) None of these
3 4 6

We have tan α = m and tan β = 1
m+1 2m + 1

m1
m + 1 + 2m + 1 2m2 + m + m + 1 tanα + tan β
tan(α + β) = m 1 = 2m2 + m + 2m + 1 − m  tan(α + β) = 1 − tanα tan β 
(m + (2m +  
1 − 1) . 1) 

= 2m 2 + 2m + 1 =1⇒ tan(α + β ) = tan π
2m 2 + 2m + 1 4

Hence α + β = π
4

Trick : As α +β is independent of m, therefore put m = 1, then tan α = 1 and tan β = 1 .
2 3

11
2+3
Therefore, tan(α + β ) = 1 = 1 , Hence α +β = π (Also check for other values of m)
6 4
1 −

If tanθ − cotθ = a and sinθ + cosθ = b , then (b 2 − 1)2 (a 2 + 4) = [WB JEE 1979]

(a) 2 (b) − 4 (c) ± 4 (d) 4

Given that tanθ − cotθ = a …….(i) and sinθ + cosθ = b …….(ii)
Now, (b 2 − 1)2 (a 2 + 4) = {(sinθ + cosθ )2 − 1}2 {tanθ − cotθ )2 + 4}

= [1 + sin 2θ − 1]2[tan2 θ + cot 2 θ − 2 + 4] = sin 2 2θ (cosec 2θ + sec 2 θ ) = 4 sin 2 θ cos 2 θ  1 + 1 = 4
 sin 2 cos2 θ 
θ

Trick : Obviously the value of expression (b 2 − 1)2 (a 2 + 4) is independent of θ , therefore put any suitable value of θ.
Let θ = 45o , we get a = 0 , b = 2 so that [( 2)2 − 1]2(02 + 4) = 4 .

Example: 34 If sin B = 1 sin(2 A + B) , then tan(A + B) =
5 tan A
Solution: (c)
(a) 5 (b) 2 (c) 3 (d) 3
Example: 35 3 3 2 5

Solution: (c) sin(2A + B) = 5 by componendo and Dividendo. sin(2 A + B) + sin B = 5 + 1
Example: 36 sin B 1 sin(2 A + B) − sin B 5 − 1
Solution: (d)
2 sin(A + B). cos A = 6 ⇒ tan(A + B) = 3 .
Example: 37 2 cos(A + B). sin A 4 tan A 2
Solution: (b)
Example: 38 sin 70o + cos 40o = [Karnataka CET 1986; MP PET 1999]
cos 70o + sin 40o
Solution: (d)
Example: 39 (a) 1 (b) 1 (c) 3 (d) 1
Solution: (b) 3 2
Example: 40
sin 70o + cos 40o = sin 70o + sin 50o 2 sin 60o cos10o = sin 60o 3 . 2 = 3.
cos 70o + sin 40o sin 20o + sin 40o = 2 sin 30o cos(−10o ) sin 30o = 2 1

sin 47o + sin 61o − sin 11o − sin 25o = [EAMCET 2003; MP PET 2001]

(a) sin 36o (b) sin 7o (c) cos 36o (d) cos 7 o

sin 47o + sin 61o − (sin 11o + sin 25o) = 2 sin 54o. cos 7o − 2 sin 18o cos 7o

= 2 cos 7o(sin 54o − sin 18o) = 2 cos 7o.2 cos 36o. sin 18o = 4. cos 7o. 5+ 1 . 5− 1 = cos 7o .
4 4

cos 10o − sin 10o = [MP PET 2002]
cos 10o + sin 10o

(a) tan 55o (b) cot 55o (c) − tan 35o (d) − cot 35o

cos10o − sin10o = 1− tan10o = tan 35o = tan(90o − 35o ) = cot 55o .
cos10o + sin10o 1+ tan10o

If tan(A + B) = p and tan(A − B) = q then the value of tan 2A = [MP PET 2002]

(a) p+q (b) p−q (c) 1 + pq (d) p+q
p−q 1 + pq 1− p 1 − pq

2A = {(A + B) + (A − B)} ⇒ tan 2A = tan(A + B) + tan(A − B) ⇒ tan 2A = p+q
1 − tan(A + B). tan(A − B) 1 − pq

sin 163o cos 347o + sin 73o sin 167o = [MP PET 2000]

(a) 0 (b) 1 (c) 1 (d) None of these
2

sin(90o + 73o ).cos(360o − 13o ) + sin 73o.sin(180o − 13o ) = cos 73o. cos 13o + sin 73o. sin 13o = cos(73o − 13o ) = cos 60o = 1 .
2

The value of cot 70o + 4 cos 70o is [Orissa JEE 2003]

(a) 1 (b) 3 (c) 2 3 (d) 1
3 2

Solution: (b) cot 70o + 4 cos 70o = cos 70o + 4 sin 70o. cos 70o = cos 70o + 2 sin140o
Example: 41 sin 70o sin 70o
Solution: (b)
Example: 42 = cos 70o + 2 sin(180 − 40o ) = sin 20o + sin 40o + sin 40o
Solution: (b) sin 70o sin 70o

= 2 sin 30o cos 10o + sin 40o = sin 80o + sin 40o = 2 sin 60o cos 20o = 3.
sin 70o sin 70o sin 70o

If tan α = (1 + 2−x )−1, tan β = (1 + 2x+1)−1, then α + β equals [AMU 2002]

(a) π (b) π (c) π (d) π
6 4 3 2

1 + 1
1 + 2x+1
tanα + tan β 1+ 1
1 − tanα tan β 2x
tan(α + β) = ⇒ tan(α + β) = 1 1
1 2
1 − . 1 + x +1
+ 2x
1

⇒ tan(α + β) = 1+ 2x + 2.2x+ x + 2x + 1 2x ⇒ tan(α + β ) = 1 = tan π ⇒ α + β = π .
2x + 2.2x + 2.2x+ x − 4 4

The value of tan 70o − tan 20o = [Karnataka CET 2003]
tan 50o

(a) 1 (b) 2 (c) 3 (d) 0

sin 70o sin 20o = sin 70o cos 20o − cos 70o sin 20o = 2 × sin(70 o − 20 o ) cos 50 o 2 sin 50o. cos 50o
cos 70o − cos 20o cos 70o. cos 20o 2 cos 70 o. cos 20 o. sin 50o = 2 cos 70o cos 20o. sin 50o
sin 50o
sin 50o

cos 50o cos 50o

= 2 cos 50o 2 cos 50o =2.
cos 90o + cos 50o = 0 + cos 50o

Trigonometric Ratio of Multiple of an Angle.

(1) sin 2A = 2 sin A cos A = 2 tan A
1 + tan2 A

(2) cos 2A = 2 cos2 A −1 = 1− 2 sin2 A = cos2 A − sin2 A = 1− tan2 A ; where A ≠ (2n + 1) π .
1+ tan2 A 4

(3) tan 2A = 1 2 tan A (4) sin 3A = 3 sin A − 4 sin3 A = 4 sin(60o − A). sin A. sin(60o + A)
− tan2 A

(5) cos 3A = 4 cos3 A − 3 cos A = 4 cos(60o − A). cos A. cos(60o + A)

(6) tan 3A = 3 tan A − tan3 A = tan(60o − A). tan A. tan(60o + A) , where A ≠ nπ + π / 6
1− 3 tan2 A

(7) sin 4θ = 4 sinθ . cos3 θ − 4 cosθ sin3 θ (8) cos 4θ = 8 cos4 θ − 8 cos2 θ + 1

(9) tan 4θ = 4 tan θ − 4 tan3 θ (10) sin 5 A = 16 sin5 A − 20 sin3 A + 5 sin A
1− 6 tan2 θ + tan4 θ

(11) cos 5 A = 16 cos5 A − 20 cos3 A + 5 cos A

Trigonometric Ratio of Sub-multiple of an Angle.

(1) sin A + cos A = 1 + sin A or sin A + cos A = ± 1 + sin A i.e., +, If 2nπ − π / 4 ≤ A / 2 ≤ 2nπ + 3π
2 2 2 2 4
 −, otherwise

(2) sin A − cos A = 1 − sin A or (sin A − cos A ) = ± 1 − sin A i.e., +, If 2nπ +π /4 ≤ A/ 2 ≤ 2nπ + 5π
2 2 2 2  4
 −, otherwise

(3) (i) tan A = ± tan2 A + 1 − 1 = ± 1 − cos A = 1 − cos A , where A ≠ (2n + 1)π
2 tan A 1 + cos A sin A

(ii) cot A = ± 1 + cos A = 1 + cos A , where A≠ 2nπ
2 1 − cos A sin A

The ambiguities of signs are removed by locating the quadrants in which A lies or you can follow the following
2

figure,

π

3π sin A + 2 A is +ve π
4 cos 2 4
2

sin A – cos A is+ve
2 2

sin A A is –ve sin A + cos A is +ve
2 + cos 2 2 2

π

sin A – cos A is +ve sin A – cos A is –ve
2 2 2 2

sin A + cos A
2 2 is –ve

5π sin A – cos A is –ve 7π
4 2 2 4

3π

2

(4) tan2 A = 1 − cos A ; where A ≠ (2n + 1)π (5) cot 2 A = 1 + cos A ; where A≠ 2nπ
2 1 + cos A 2 1 − cos A

Important Tips

o Any formula that gives the value of sin A in terms of sin A shall also give the value of sine of nπ + (−1)n A .
2 2

 Any formula that gives the value of cos A in terms of cos A shall also give the value of cos of 2nπ ± A .
2 2

 Any formula that gives the value of tan A in terms of tan A shall also give the value of tan of nπ ± A .
2 2

Example: 43 If sin α = −3 where π <α < 3π , then cos α equal to [MP PET 1998]
5 2 2
Solution: (d) (d) −3
Example: 44 (a) 1 (b) − 1 (c) 3 10
Solution: (c) 10 10 10
Example: 45
3π π α 3π cos α −ve cos α 4 cos α 1 + cos α 1+ 4 9 −3 .
2 2 2 4 2 5 2 2 5 10 = 10
π < α < ⇒ < < ⇒ = ⇒ ∴ = ⇒ = =− 2 =−

2 sin 2 β + 4 cos(α + β ) sinα sin β + cos 2(α + β ) equal to [UPSEAT 1993; IIT 1977]

(a) sin 2α (b) cos 2β (c) cos 2α (d) sin 2β

Since 2 cos(α + β ) = 2 cos2(α + β ) − 1, 2 sin 2 β = 1 − cos 2β = − cos 2β + 2 cos(α + β )[2 sinα sin β + cos(α + β )]
= − cos 2β + 2cos(α + β ).cos(α − β ) = − cos 2β + cos 2α + cos 2β = cos 2α .

cot2 15o − 1 = [MP PET 1998]
cot2 15o + 1

(a) 1 (b) 3 (c) 33 (d) 3
2 2 4

1 − 1 2

1−  3  1 −  3 −12
1 + 1   
1 − 3 1  [ 3 + 1]2 − [ 3 − 1]2 43 3
Solution: (b) 1 − tan2 15 1 − [tan(45o − 30o )]2 3  =  + = [ 3 + 1]2 + [ 3 − 1]2 = 8 = 2
1 + tan2 15 1 + [tan(45o − 30o )]2
Example: 46 = = 1 2  3 −12
Solution: (d) 
3  1 + 3 + 1
1 
1+   

1 + 3 

Trick : cos 2θ = 1 − tan2 θ ⇒ 1 − tan2 15o = cos 30o = 3 .
1 + tan2 θ 1 + tan2 15o 2

If sin 6θ = 32cos5 θ .sinθ − 32cos3 θ sinθ + 3x, then x = [EAMCET 2003]

(a) cosθ (b) cos 2θ (c) sinθ (d) sin 2θ

sin 6θ = 2 sin 3θ .cos 3θ = 2[3 sinθ − 4 sin 3 θ ][4 cos 3 θ − 3 cosθ ]

= 24 sinθ .cosθ (sin2 θ + cos2 θ ) − 18 sinθ cosθ − 32 sin3 θ cos3 θ = 32 cos5 θ . sinθ − 32 cos3 θ . sinθ + 3 sin 2θ
On comparing, x = sin 2θ
Trick : Put θ = 0o , then x = 0 . So, option (c) and (d) are correct.

Now put θ = 30o , then x = 3 . Therefore, Only option (d) is correct.
2

Example: 47 If x + 1 = 2cosθ , then x 6 + x −6 = [Karnataka CET 2003]
Solution: (b) x

Example: 48 (a) 2cos 6θ (b) 2cos12θ (c) 2cos 3θ (d) 2 sin 3θ
Solution: (c)
Example: 49 Given, x + 1 = 2cosθ ........(i)
Solution: (b) x

On squaring both sides we get, x + 1 + 2 = 4 cos 2 θ ⇒ x + 1 = 4 cos 2 θ − 2
x x

⇒ x + 1 = 2(2cos 2 θ − 1) = 2cos 2θ ........(ii)
x

Again squaring both sides,

x2 + 1 +2= 4 cos 2 2θ ⇒ x2 + 1 = 4 cos 2 2θ − 2 = 2(2cos 2 2θ − 1) ⇒ x2 + 1 = 2cos 4θ ......(iii)
x2 x2 x2

Now taking cube of both sides;  x 2 + 1 3 = (2 cos 4θ )3 ⇒ x6 + 1 + 3x 2. 1  x 2 + 1  = 8 cos 3 4θ
 x2  x6 x2  x2 

⇒ x6 + 1 + 3(2 cos 4θ ) = 8 cos3 4θ ⇒ x6 + 1 = 8 cos 3 4θ − 6 cos 4θ
x6 x6

⇒ x6 + 1 = 2(4 cos 3 4θ − 3 cos 4θ ) = 2 cos 3(4θ ) = 2 cos 12θ .
x6

For A = 133o, 2 cos A is equal to [DCE 2001]
2

(a) − 1 + sin A − 1 − sin A

(b) − 1 + sin A + 1 − sin A

(c) 1 + sin A − 1 − sin A

(d) 1 + sin A + 1 − sin A

For A = 133o , A = 66.5o ⇒ sin A > cos A > 0
2 2 2

Hence, 1+ sin A = sin A + cos A ......(i) and 1− sin A = sin A − cos A ......(ii)
2 2 2 2

Subtract (ii) from (i) we get, 2 cos A = 1 + sin A − 1 − sin A .
2

If 2 tan A = 3 tan B, then sin 2B is equal to [AMU 2001]
5 − cos 2B
(d) tan(A + 2B)
(a) tanA – tanB (b) tan(A − B) (c) tan(A + B)

2 tan A = 3 tan B ⇒ tan A = 3 tan B = 3 t (Let tan B = t ) ⇒ sin 2B = 1 2t 2 , cos 2B = 1 − t 2
2 2 +t 1 + t 2

sin 2B  1 2t 2  2t t = tan(A − B) .
5 − cos 2B  +t  4 + 6t 2 2 + 3t 2
∴ = = =
 1 t 2 
5 − 1 − t 2
+

Example: 50 If 90o < A < 180o and sin A = 4 , then tan A is equal to [AMU 2001]
Solution: (d) 5 2

Example: 51 (a) 1 (b) 3 (c) 3 (d) 2
Solution: (a) 2 5 2

Example: 52 sin A = 4 ⇒ tan A = − 4 , (90o < A < 180o)
Solution: (a) 5 3
Example: 53
tan A 2 tan A , Let tan A =P
2 2
= A
tan2 2
1−

⇒ −4 = 2P ⇒ 4P 2 − 6P − 4 = 0 ⇒ P = − 1 , 2 ⇒ P 1 (impossible)
3 1− P2 2 =−2

So, P=2 i.e., tan A = 2.
2

If tan α = 1 and sin β = 1 , then tan(α + 2β ) is equal to [AMU 2001]
7 10

(a) 1 (b) 0 (c) 1 (d) 3
2 4

1 1 1 2 3
7 10 3 =4
tan α = , sin β = ⇒ tan β = ⇒ tan 2β = 3 1
9
1−

tan(α 2β ) 1 + 3 4 + 21 1
7 4 25
∴ + = 3 = =
28
1−

If tan θ = t, then 1 − t2 is equal to [Kerala (Engg.) 2002]
2 1 + t2
(d) cos 2θ
(a) cosθ (b) sinθ (c) sec θ
[Haryana CEE 1998; IIT 1992, 97]
1− t2 1 − tan 2 θ
1+ t2 2 (d) 5 and 6
= (∴ tan θ = t) = cos(2.θ2) = cosθ .
tan2 θ 2
1 + 2

The value of tan x when ever defined never lie between
tan 3x

(a) 1 and 3 (b) 1 and 4 (c) 1 and 5
3 4 5

Solution: (a) Let, y= tan x = tan x
tan 3x 3 tan x − tan3 x
Example: 54
Solution: (a) 1 − 3 tan2 x
Example: 55
Solution: (b) y 1 − 3 tan2 x 1 − tan2 x
3 − tan2 x 3
Example: 56 = = 1
Solution: (b) 3 tan2
1 − x

Hence, y should never lie between 1 and 3 whenever defined.
3

If tanθ = t , then tan 2θ + sec 2θ equal to [MP PET 1999]
[MP PET 2000]
(a) 1+ t (b) 1−t (c) 2t (d) 2t
1−t 1+ t 1−t 1+ t

tan 2θ + sec 2θ = 2 tanθ + 1 + tan2 θ
1 − tan2 θ 1 − tan2 θ

Given tanθ = t ⇒ ∴ tan 2θ + sec 2θ = 2t + 1 + t 2 = 2t + 1 + t 2 = (t + 1)2 = 1+ t .
1− t2 1 − t 2 1− t2 1− t2 1−t

If sin 2θ + sin 2φ = 1 and cos 2θ + cos 2φ = 3 , then cos 2 (θ − φ) equal to
2 2

(a) 3 (b) 5 (c) 3 (d) 5
8 8 4 4

Given, sin 2θ + sin 2φ = 1 .......(i) and cos 2θ + cos 2φ = 3 .......(ii)
2 2

Squaring and adding, ∴(sin2 2θ + cos2 2θ ) + (sin2 2φ + cos2 2φ) + 2[sin 2θ . sin 2φ + cos 2θ . cos 2φ ] = 1 + 9
4 4

⇒ cos 2θ . cos 2φ + sin 2θ . sin 2φ = 1 ⇒ cos(2θ − 2φ) = 1 ⇒ cos2(θ − φ) = 5 .
4 4 8

If tan x = b , then a+b + a−b equal to [MP PET 1990, 2002]
a a−b a+b
(d) 2 sin x
(a) 2 sin x (b) 2 cos x (c) 2 cos x cos 2x
sin 2x cos 2x sin 2x

Given tan x = b ⇒ a+b + a−b = 1+b/a + 1−b/a = 1 + tan x + 1 − tan x = 2
a a−b a+b 1−b/a 1+b/a 1 − tan x 1 + tan x 1 − tan 2 x

Now, multiplying by 1 + tan 2 x in N'r and D'r = 2 2 = 2 cos x .
= cos 2x . sec 2 x cos 2x

1 − tan 2 x . 1 + tan 2 x
1 + tan 2 x

Maximum and Minimum Value of a cosθ + b sinθ.

Let a = r cosα ……..(i) and b = r sinα ……..(ii)

Squaring and adding (i) and (ii), then a 2 + b 2 = r 2 or, r = a 2 + b 2
∴ a sinθ + b cosθ = r(sinθ cosα + cosθ sinα) = r sin(θ + α)
But − 1 ≤ sinθ < 1 So, − 1 ≤ sin(θ + α) ≤ 1; Then − r ≤ r sin(θ + α) ≤ r

Hence, − a 2 + b 2 ≤ a sinθ + b cosθ ≤ a 2 + b 2

Then the greatest and least values of a sin θ + b cos θ are respectively a 2 + b 2 and − a 2 + b 2 .

Note :  sin 2 x + cosec 2 x ≥ 2, for every real x.

 cos 2 x + sec 2 x ≥ 2, for every real x.
 tan 2 x + cot 2 x ≥ 2 , for every real x.

Important Tips

Use of Σ (Sigma) and ∏ (Pie) notation
 sin(A + B + C) = Σ sin A cos B cos C − Π sin A , cos(A + B + C) = Π cos A − Σ cos A sin B sin C ,

tan(A + B + C) = Σ tan A − Π tan A . ( Σ denotes summation)
1 − Σ tan A tan B ( ∏ denotes product)

 sin α + sin(α + β ) + sin(α + 2β ) + .........n terms

n sin A + n −1 B  sin nB
r∑=1 sin(A +  2  2
= sin[α + (n − 1)β / 2] sin[nβ / 2] or r − 1B) = .
sin(β / 2)
sin B
2

n cos A + n − 1 B  sin nB
r∑=1 cos(A + r − 1B) =  2  2
 cos α + cos(α + β ) + cos(α + 2β ) + .........n terms = cos[α + (n − 1)β / 2] sin[nβ / 2] or .
sin[β / 2]
sin B
2

 sin A / 2 ± cos A / 2 = 2 sin[π / 4 ± A] = 2 cos[A  π / 4].

 cos α + cos β + cos γ + cos(α + β +γ)= 4 cos α +β cos β + γ cos γ +α .
2 2 2

 sin α + sin β + sin γ − sin(α +β +γ)= 4 sin α + β sin β +γ sin γ +α .
2 2 2

• tan α + 2 tan 2α + 4 tan 4α + 8 cot 8α = cot α .

Example: 57 If x = y cos 2π = z cos 4π , then xy + yz + zx = [EAMCET 1994]
3 3

(a) −1 (b) 0 (c) 1 (d) 2

Solution: (b) We have x = y = z = λ (say)
1 −2 −2

∴ x = λ , y = −2λ, z = −2λ ; ∴ xy + yz + zx = −2λ2 + 4λ2 − 2λ2 = 0

Example: 58 sec 8 A − 1 equal to [MP PET 1995]
sec 4 A − 1

(a) tan 2A (b) tan 8 A (c) cot 8 A (d) None of these
tan 8 A tan 2A cot 2A

Solution: (b) 1 − cos 8A . cos 4 A = 2 sin 2 4A . cos 4A = 2 sin 4 . A cos 4 A.sin 4 A = sin 8 A.2 sin 2A.cos 2A = tan 8A .
cos 8 A − cos 4A cos 8A 2 sin 2 2A cos 8 A.2 sin2 2A cos 8 A.2 sin2 2A tan 2A
1

Example: 59 If tanθ = a , then sinθ + cosθ equal to [WB JEE 1986]
b cos8 θ sin8 θ

(a) ± (a2 + b2)4 a + b (b) ± (a2 + b2)4 a − b
a2 + b2  b8 a8  a2 + b2  b8 a8 

(c) ± (a2 − b2)4 a + b (d) ± (a2 − b2)4 a − b
a2 + b2  b8 a8  a2 + b2  b8 a8 

Solution: (a) Given , tanθ = a / b ⇒ cos 2θ = 1− tan2 θ = b2 − a2
1+ tan2 θ b2 + a2

sinθ = ± a ; cosθ = ± b
a2 + b2 a2 + b2

 a   b  = a(a2 + b2)4 b(a2 + b2)4
 a2 + b2   a2 + b2  b8 (a2 + b2)1 / 2 a8 (a2 + b2)1 / 2
∴ sinθ + cosθ =  b 8 +  a 8 + = ± (a2 + b2)4 a + b .
cos8 θ sin8 θ a2 + b2  b8
a8 

 a2 + b2   a2 + b2 

Example: 60 The minimum value of 3 cos x + 4 sin x + 5 = [UPSEAT 1991]
Solution: (d)
Example: 61 (a) 5 (b) 9 (c) 7 (d) 0
Solution: (b)
Example: 62 Minimum value of 3 cos x + 4 sin x = − 3 2 + 4 2 = − 5
Solution: (b)
Minimum value of 3 cos x + 4 sin x + 5 = −5 + 5 = 0 . [UPSEAT 1975]
Example: 63 The greatest and least value of sin x cos x are
Solution: (b) (d) 2,− 2
Example: 64 (a) 1, − 1 (b) 1 , − 1 (c) 1 , − 1
Solution: (d) 2 2 4 4 [UPSEAT 1977, 83; RPET 1995]

1 [2 sin x cos x] = 1 sin 2x ; −1 ≤ sin 2x ≤ 1 ; −1 ≤ sin 2x ≤ 1 . (d) θ = 90o
2 2 2 2 2

The value of sinθ + cosθ will be greatest when

(a) θ = 30o (b) θ = 45o (c) θ = 60o

Let f(x) = sinθ + cosθ = 2 sin(θ + π )
4

−1 ≤ sin(θ + π ) ≤ 1 ⇒ − 2≤ 2 sin(θ + π ) ≤ 2
4 4

If f(x) is maximum then,

sin(θ + π ) = 1 = sin π ⇒θ = π ⇒ θ + π = π ⇒θ = π .
4 2 4 4 2 4

The maximum value of sin 2 x + 3 cos 2 x is [Karnataka CET 2003]

(a) 3 (b) 4 (c) 5 (d) 7

f(x) = 4 sin 2 x + 3 cos 2 x = sin 2 x + 3 and 0 ≤|sin x|≤ 1

∴ Maximum value of sin 2 x + 3 cos 2 x is 4.

If A = cos 2 θ + sin 4 θ , then for all values of θ [UPSEAT 2001; IIT 1980; Roorkee 1992; EAMCET 1994]

(a) 1 ≤ A ≤ 2 (b) 13 ≤ A ≤1 (c) 3 ≤ A ≤ 13 (d) 3 ≤ A ≤1
16 4 16 4

A = cos 2 x + sin 4 x ⇒ cos 2 θ + sin 2 θ sin 2 θ

⇒ A ≤ cos 2 θ + sin 2 θ [ sin 2 θ ≤ 1 ] ⇒ A ≤ 1

Again A = cos 2 θ + sin 4 θ = (1 − sin 2 θ ) + sin 4 θ

A =  sin2 θ − 1 3 + 3 ≥ 3
 2  4 4

Hence, 3 ≤ A ≤1.
4

Example: 65 The value of 5 cosθ + 3 cos(θ + π ) + 3 lies between
Solution: (d) 3

Example: 66 (a) − 4 and 4 (b) − 4 and 6 (c) − 4 and 8 (d) − 4 and 10
Solution: (d)
Example: 67 5 cosθ + 3 cos(θ + π ) + 3 = 5 cosθ + 3[cosθ cos π − sinθ . sin π ] + 3
Soluton: (b) 3 3 3

= [5 cosθ + 3 cosθ − 33 sinθ ] + 3 =  13 cosθ − 33 sinθ  + 3
2 2  2 2 

−  13 2 +  33 2 ≤  13 cosθ − 3 3 sinθ  ≤  13 2 +  33 2
 2   2   2 2   2   2 

−7 ≤  13 cos θ − 33 sin θ  ≤ +7
 2 2 

∴ − 7 + 3 ≤  13 cos θ − 33 sin θ  + 3 ≤ 7 + 3 ⇒ − 4 ≤  13 cos θ − 33 sin θ  + 3 ≤ 10
 2 2   2 2 

So, the value lies between – 4 and 10.

sin π . sin 3π . sin 5π . sin 7π . sin 9π . sin 11π . sin 13π is equal to [IIT 1991; UPSEAT 1992]
14 14 14 14 14 14 14

(a) 1 (b) 1 (c) 1 (d) 1
8 16 32 64

sin π . sin 3π . sin 5π . sin 7π . sin 9π . sin 11π . sin 13π
14 14 14 14 14 14 14

= sin π . sin 3π . sin 5π × 1 × sinπ − 5π . sinπ − 3π . sinπ − π  = sin π . sin 3π . sin 5π . sin 7π 2 = 1 .
14 14 14  14  14  14  14 14 14 14  64

If sinθ + sinφ = a and cosθ + cosφ =b then tan θ −φ equal to [MP PET 1993]
2

(a) a2 + b2 (b) 4 − a2 − b2 (c) a2 + b2 (d) 4 + a2 + b2
4 − a2 − b2 a2 + b2 4 + a2 + b2 a2 + b2

Given that, sinθ + sinφ = a ........(i) and cosθ + cosφ = b ......(ii)

Squaring, sin 2 θ + sin 2 φ + 2 sinθ sinφ = a 2 and cos2 θ + cos2 φ + 2 cos θ cos φ = b2

Adding, 2 + 2(sinθ sinφ + cosθ .cosφ) = a 2 + b 2

a2 + b2 − 2 1− tan2 (θ −φ) a2 + b2 − 2
2 2 2
⇒ 2 cos(θ − φ) = a2 + b2 − 2 ⇒ cos(θ −φ) = ⇒ (θ −φ) =
tan2 2
1+

⇒ (a2 + b2) + (a2 + b2) tan2 θ −φ − 2 − 2 tan2 θ −φ = 2 − 2 tan2 θ −φ
2 2 2

⇒ 4 − a2 − b2 = tan2 (θ −φ) ⇒ tan (θ −φ) = 4 − a2 − b2
a2 + b2 2 2 a2 + b2

Trick : Put θ = π ,φ = 0o, then a =1= b
2

∴ tan θ −φ = 1, which is given by (a) and (b).
2

Again putting θ = π =φ , we get tan θ −φ =0 , which is given by (b).
4 2

Example: 68 The maximum value of 3 cosθ + 4 sinθ equal to [MP PET 2002; UPSEAT 1990]
Solution: (c)
(a) 3 (b) 4 (c) 5 (d) None of these

Maximum value of 3 cosθ + 4 sinθ is 32 + 4 2 = 5 .

Conditional Trigonometrical Identitites.
We have certain trigonometric identities. Like, sin2 θ + cos2 θ = 1 and 1 + tan2 θ = sec2 θ etc.

Such identities are identities in the sense that they hold for all value of the angles which satisfy the given
condition among them and they are called conditional identities.

If A, B, C denote the anlges of a triangle ABC, then the relation A + B + C = π enables us to establish many
important identities involving trigonometric ratios of these angles.

(1) If A + B + C = π, then A + B = π – C, B + C = π – A and C + A = π – B.

(2) If A + B + C = π, then sin(A + B) = sin(π − C) = sin C

Similarly, sin(B + C) = sin(π − A) = sin A and sin(C + A) = sin(π − B) = sin B

(3) If A + B + C = π , then cos(A + B) = cos(π − C) = − cos C

Similarly, cos(B + C) = cos(π − A) = − cos A and cos(C + A) = cos(π − B) = − cos B

(4) If A + B + C = π, then tan(A + B) = tan(π − C) = − tan C

Similarly, tan(B + C) = tan(π − A) = − tan A and tan(C + A) = tan(π − B) = − tan B

(5) If A + B + C = π , then A+B = π − C and B + C = π − A and C + A = π − B
2 2 2 2 2 2 2 2 2

sin A + B  = sin π − C  = cos C  , cos A + B  = cos π − C  = sin C  ,
 2   2 2   2   2   2 2   2 

tan A + B  = tan π − C  = cot C 
 2   2 2   2 

All problems on conditional identities are broadly divided into the following three types

1. Identities involving sine and cosine of the multiple or sub-multiple of the angles involved
Working Method

Step (i) : Use C ± D formulae.

Step (ii) : Use the given relation (A + B + C = π) in the expression obtained in step-(i) such that a factor can
be taken common after using multiple angles formulae in the remaining term.

Step (iii) : Take the common factor outside.

Step (iv) : Again use the given relation (A + B + C = π) within the bracket in such a manner so that we can
apply C ± D formulae.

Step (v) : Find the result according to the given options.

2. Identities involving squares of sine and cosine of multiple or sub-multiples of the angles
involved

Working Method

Step (i) : Arrange the terms of the identity such that either sin2 A − sin2 B = sin(A + B). sin(A − B) or
cos2 A − sin2 B = cos(A + B). cos(A − B) can be used.

Step (ii) : Take the common factor outside.
Step (iii) : Use the given relation (A + B + C = π ) within the bracket in such a manner so that we can apply
C ± D formulae.
Step (iv) : Find the result according to the given options.
3. Identities for tangent and cotangent of the angles
Working Method

Step (i) : Express the sum of the two angles in terms of third angle by using the given relation (A + B + C = π ).
Step (ii) : Taking tangent or cotangent of the angles of both the sides.
Step (iii) : Use sum and difference formulae in the left hand side.
Step (iv) : Use cross multiplication in the expression obtained in the step (iii).
Step (v) : Arrange the terms as per the result required.

Example: 69 If A + B + C = π , then cos 2 A + cos 2 B − cos 2 C equal to

(a) 1 − 2 sin A sin B cos C (b) 1 − 2 cos A cos B sin C

(c) 1 + 2 sin A sin B cos C (d) 1 + 2 cos A cos B sin C

Solution: (a) cos2 A + cos2 B − cos2 C = cos2 A + (1 − sin2 B) − cos2 C
Example: 70
Solution: (a) = 1 + [cos 2 A − sin 2 B] − cos 2 C = 1 + cos(A + B)cos(A − B) − cos 2 C

Example: 71 = 1 + cos(π − C)cos(A − B) − cos 2 C = 1 − cos C[cos(A − B) + cos C]
Solution: (a)
Example: 72 = 1 − cos C[cos(A − B) + cos{π − (A + B)}] = 1 − cos C[cos(A − B) − cos(A + B)]
Solution: (c)
= 1 − cos C[2 sin A sin B] = 1 − 2 sin A sin B cos C .

sin 2A + sin 2B − sin 2C equal to
sin A + sin B − sin C

(a) cos A cos B sin C (b) sin A sin B cos C (c) − cos A cos B sin C (d) − sin A sin B cos C
cos A
sin A sin B cos C cos A cos B sin C sin A sin B cos C cos B sin C
2 2 2 2 2 2 2 2 2 2 2 2

(sin 2A + sin 2B) − sin 2C = 2 sin(A + B)cos(A − B) − sin 2C = 2 sin C cos(A − B) − 2 sin C cos C
(sin A + sin B) − sin C 2 sin A
+ B  cos A − B  − sin C 2 sin π − C  cos A − B  − 2 sin C cos C
 2   2   2  2  2 2

2 sin C[cos(A − B) − cos C]   sin C = 2 sin C cos C 
 2= 2 2 
=  A B  = cos (A B) 
C cos A B  cos A B  sin C / sin π + + 
2 cos 2 2 − 2  −  2 + 2   2 − 2  2

= 2 sin C[cos(A − B) + cos(A + B)] = 2 sin C[2 cos A cos B] = cos A cos B sin C .

2 cos C cos A − B  − cos A + B  2 cos C 2 sin A sin B sin A sin B cos C
2  2 2   2 2  2 2 2  2 2 2

Trick :  sin 2A + sin 2B − sin 2C = 4 cos A cos B sin C

and sin A + sin B− sin C = 4 sin A sin B cos C ⇒ sin 2A + sin 2B − sin 2C = cos A cos B sin C .
2 2 2 sin A + sin B − sin C
sin A sin B cos C
2 2 2

If α + β + γ = 2π , then [IIT 1979]

(a) tan α + tan β + tan γ = tan α tan β tan γ (b) tan α tan β + tan β tan γ + tan γ tan α = 1
2 2 2 2 2 2 2 2 2 2 2 2

(c) tan α + tan β + tan γ = − tan α tan β tan γ (d) None of these
2 2 2 2 2 2

We have α + β + γ = 2π ⇒ α + β + γ = π ⇒ tan α + β + γ  = tan π =0
2 2 2  2 2 2 

⇒ tan α + tan β + tan γ − tan α . tan β . tan γ = 0 ⇒ tan α + tan β + tan γ = tan α . tan β . tan γ
2 2 2 2 2 2 2 2 2 2 2 2

If A + B + C = π , then cos 2A + cos 2B + cos 2C equal to [EAMCET 1982]

(a) 1 + 4 cos A cos B sin C (b) −1 + 4 sin A sin B cos C (c) −1 − 4 cos A cos B cos C (d) None of these

cos 2A + cos 2B + cos 2C = 2 cos(A + B).cos(A − B) + (2 cos 2 C − 1) = − 1 − 2 cos C.cos(A − B) + 2 cos 2 C

= −1 − 2cos c[cos(A − B) + cos(A + B)] = −1 − 4 cos A.cos B.cos C

Example: 73 If A + B + C = 180o , then sin 2A + sin 2B + sin 2C equal to
Solution: (b) cos A + cos B + cos C − 1

Example: 74 (a) 8 sin A sin B sin C (b) 8 cos A cos B cos C (c) 8 sin A cos B cos C (d) 8 cos A sin B sin C
Solutio: (b) 2 2 2 2 2 2 2 2 2 2 2 2

Example: 75 sin 2A + sin 2B + sin 2C = 2 sin(A + B). cos(A − B) + 2 sin C cos C = 2 sin C cos(A − B) + 2 cos C sin C
Solution: (c) cos A + cos B + cos C − 1
2 cos A+ B cos A− B + 1 − 2 sin2 C −1 2 sin C cos A − B − 2 sin 2 C
Example: 76 2 2 2 2 2 2
Solution: (d)
= 2 sin C[cos(A − B) + cos(A + B)] = 4 sin A sin B sin C
B
2 sin C cos (A − B) − cos (A + B)  4 sin A sin sin C
2 2 2  2 2 2

4 × 2 sin A × cos A × 2 sin B cos B × 2 sin C cos C A B C .
2 2 2 2 2 2 2 2 2
= A B C = 8 cos cos cos
2 2 2
4 sin sin sin

If A + B + C = 180o , then the value of (cot B + cot C)(cot C + cot A)(cot A + cot B) will be [UPSEAT 1999]

(a) sec A sec B sec C (b) cosec A cosec B cosec C (c) tan A tan B tan C (d) 1

cot B + cot C = sin C cos B + sin B cos C = sin(B + C) = sin(180o − A) = sin A
sin B. sin C sin B. sin C sin B. sin C sin B. sin C

Similarly, cot C + cot A = sin B A and cot A+ cot B = sin C
sin C. sin sin A sin B

Therefore, (cot B + cot C)(cot C + cot A)(cot A + cot B)

= sin A C . sin B A . sin C B = cosecA.cosecB.cosecC .
sin B. sin sin C. sin sin A sin

If A + B + C = 180o , then the value of cot A + cot B + cot C will be [UPSEAT 1999]
2 2 2

(a) 2 cot A cot B cot C (b) 4 cot A cot B cot C (c) cot A cot B cot C (d) 8 cot A cot B cot C
2 2 2 2 2 2 2 2 2 2 2 2

A + B + C = 180o ∴ A + B = 90o − C
2 2 2

∴ cot A B  cot 90o C  or cot A . cot B −1 tan C 1
 2 2   2  2 2 2
+ = − B A = = C
2 2 2
cot + cot cot

or  cot A . cot B − 1 cot C = cot B + cot A ; cot A . cot B . cot C = cot C + cot B + cot A
 2 2  2 2 2 2 2 2 2 2 2

If A, B, C are angles of a triangle, then sin 2A + sin 2B − sin 2C is equal to [MP PET 2003]

(a) 4 sin A cos B cos C (b) 4 cos A (c) 4 sin A cos A (d) 4 cos A cos B sin C

sin 2A + sin 2B − sin 2C = 2 sin A cos A + 2 cos(B + C)sin(B − C)

[ A + B + C = π , B + C = π − A, cos(B + C) = cos(π − A), cos(B + C) = − cos A, sin(B + C) = sin A]

= 2 cos A[sin A − sin(B − C)] = 2 cos A[sin(B + C) − sin(B − C)] = 2 cos A.2 cos B. sin C = 4 cos A.cos B. sin C

Trick: First put A = B = C = 60o , for these values. Options (a) and (b) satisfies the condition.

Now put A = B = 45o and C = 90o , then only (d) satisfies.
Hence (d) is the answer.

Example: 77 In any triangle ABC sin 2 A + sin 2 B + sin 2 C is equal to [MP PET 2003]
Solution: (c) 2 2 2

(a) 1 − 2 cos A cos B cos C (b) 1 − 2 sin A cos B cos C
2 2 2 2 2 2

(c) 1 − 2 sin A sin B sin C (d) 1− 2 cos A cos B sin C
2 2 2 2 2 2

Trick: For A = B = C = 60o only option (c) satisfies the condition.

Important Tips

 Method of componendo and dividendo

If p = a , then by componendo and dividendo
q b

We can write p+q = a+b or q+ p = b+a or p−q = a−b or q− p = b−a .
p−q a−b q−p b−a p+q a+b q+ p b+a

Example: 78 If tan β = cosθ . tanα then tan2 θ equal to
2

(a) sin(α − β ) (b) sin(α + β ) (c) cos(α − β ) (d) cos(α + β )
sin(α + β ) sin(α − β ) cos(α + β ) cos(α − β )

Solution: (a) The given relation is tanα = 1
tan β cosθ

Applying componendo and dividendo rule, then

tanα − tan β 1 − cosθ sin(α − β) 2 sin2 θ sin(α − β) tan2 θ .
tanα + tan β 1 + cosθ sin(α + β) 2 sin(α + β) 2
⇒ = ⇒ = θ ⇒ =
cos2 2
2

Example: 79 If m cos(θ + α) = n cos(θ − α) , then cotθ cotα equal to

(a) m+n (b) m−n (c) m+n (d) n−m
m−n m+n n−m n+m

Solution: (c) m = cos(θ −α)
n cos(θ + α)

By componendo and dividendo rule, m+n = cos(θ − α) + cos(θ + α) ⇒ m + n = 2 cosθ cosα
m−n cos(θ − α) − cos(θ + α) m − n − 2 sinθ sinα

cot θ cos α = m+n .
n−m

Example: 80 If cosecθ = p + q , then cot π + θ  = [EAMCET 2001]
p − q  4 2 

(a) p (b) q (c) pq (d) pq
q p

Solution: (b) Given, cosecθ = p+q ⇒ 1 = p + q ,
p−q sinθ p − q

Apply componendo and dividendo, 1 + sinθ = p+q+ p−q ⇒  cos θ + sin θ 2 = p
1 − sinθ p+q− p−q  cos 2 − sin 2  q
 θ θ 
 2 2 
 

⇒ 1 + tanθ / 22 = p ⇒ tan2  π + θ  = p ⇒ cot 2  π + θ  = q
1 − tanθ / 2  q  4 2  q  4 2  p

Note : cot π + θ  = q only if cot π + θ  >0.
 4 2  p  4 2 