Opt. Maths Book for Class-9

Solution
Given equation is

x–y+1= 0

or, –x+y= 1

Dividing both sides by Y

= (–1)2 +12 = √2 P

–x + y = 1 B(0,1)
√2 √2 √2 C5
1
x.cos135°+y.sin135°= √2

comparing with x.cosα +ysinα = p

we get p= 1 or, OC= 1
√2 √2
1 2 X'
From ∆OCP, PC= op2 – oc2 = √2 X
52 – A(–1,0) O
5 Y'
= 25 –12 = 7
√2 Q

Area of ∆POQ==1221O×C√×12PQ×2×√72 [∴ PQ= 2.PC]


= 3.5sq.unit.

31. Find the equation of straight line passing through the points (4,2) and (7,–1). Prove that
the point (–1,7) lies on the line.

Solution
Here, (x1,y1) = (4,2)

( x2,y2) = (7,–1)

The equation of line is
xy22––yx11
y–y1 = (x–4) (x–x1)

or, y–2= –1–2
7–4
–3
or, y–2= 3 (x–4)

or, y–2= –x+4

⸫ x+y–6= 0, which is the required equation.

Putting (–1,7) on the equation, we get

–1+7–6=0

or, 0= 0 (true)

Hence the line passes through (–1,7).

32.(a) Find the equation of a straight line passing through the point of intersection of line
2x+3y= 17 and 9x–8y= 12 and having slope 2

Solution
Given two lines
2x+3y= 17 ............(i)

Vedanta Optional Mathematics Teacher's Guide ~ 9 101

9x–8y= 12 .............(ii)

Multiplying equation (i) by 8 and equation (ii) by 3, then adding both

16x+24y= 136
27x–24y= 36
43x= 172

∴ x= 173 = 4
43
putting value of x, in equation (i)

2x+3y= 17

or, 2×4+3y= 17

or, 8+3y= 17

3y= 9

y= 3

The point of intersection is (4,3)

The equation of line passing through (4,3) and with slope 2 is given by

y–y1 = m(x–x1)
or, y–3= 2(x–4)

or, y–3= 2x–8

⸫ 2x–y–5= 0, which is the required equation.

(b) Find the equation of a straight line passing through the point(1,2) and centroid of a
triangle formed by the points (2,1), (–5,1) and (–3,7).

Solution
vertices of triangle are (2,1), (–5,1) and (–3,7)

centroid of triangle x1 + x32 + x3 , y1 + y32 +y3 ,

= 2–5–3 , 1 + 1 +7 ,
3 3

=(–2,3)

The equation of line passing through (1,2) and centroid i.e. (–2,3) is given by

y– y1 = –3xy2–22–––21yx11(x(x–1–x) 1)
or, y–2=

or, y–2= –1 (x–1)
–3
or, 3y–6= –x+1

⸫ x+3y= 7, which is the required equation.

33. (a) Find the equation of the median of the triangle formed by the points (2,2), (2,8), and
(–6,2) drawn from the first vertex (2,2)

Solution
Let A(2,2), B(2,8) and c(–6,2) are the vertices of the triangle.

D is the mid point of BC.

102 Vedanta Optional Mathematics Teacher's Guide ~ 9

Coordinates of D are given by

x1 + x2 , y1 + y2 A (2,2)
2 2

or, 2–6 , 8+2
2 2

or, (–2,5)

Equation of Median AD,

y–y1 = xy22––xy11 (x–x1)
5–2
or, y–2= –2–2 (x–2) B (2,8) D C (–6,2)

or, y–2= –3 (x–2)
–4

or, 4y–8= –3x+6

or, 3x+4y–14= 0

34. (a) P(a,b) lies on the line 6x–y= 1 and Q(b,a) lies on the line 2x–5y= 5. Find the equation
of PQ. Find the length of PQ.

Solution
P(a,b) lies on the line 6x–y= 1

6a–b= 1 ...............(i)

Q (b,a) lies on the line 2x–5y= 5

2b–5a= 5 .............(ii)

from equation (i), 6a–b= 1

or, b= 6a–1

Now from equation (ii), put the value of b.

2b–5a= 5

2(6a–1)–5a= 5

12a–2–5a= 5

7a= 7

a= 1
from equation (i)

6×1–b= 1

or, b= 5
The points are P(1,5) and Q(5,1)

The equation of PQ is given by

y–y1 = xy22––yx11 (x–x1)

or, y–5= 1–5 (x–1)
5–1

or, y–5= –4 (x–1)
4
or, y–5= –x+1

⸫ x+y–6= 0, which is the required equation.

Vedanta Optional Mathematics Teacher's Guide ~ 9 103

(b) Let P and Q be two points on the line x–y+1= 0 such that each of them is 5 units far from
the origin. Find the coordinates of point of the line. Also find the equation of OP and OQ.

Solution
Let P(a,b) be point on the line x–y+1= 0 which is at 5 units distance from origin.

∴ a–b+1 = 0 Y
or, a= b–1 .............(i)

Distance OP= (x2– x1)2 + (y2– y1)2 P

or, 5= (a– 0)2 + (b– 0)2 x–y+1 = 0
or, 25= a²+b² C5
or, 25= (b–1)+b²

or, 25= b²–2b+1+b² X' OX
or, 2b²–2b–24= 0 5 Y'
or, b²–b–12= 0
or, b²–4b+3b–12= 0 Q
or, (b–4)(b+3)= 0

or, b=4,–3

when b=4, a=3

b=–3, a=–4

Since P has two coordinates, another coordinates must be of Q which is also at 5 unit distance.

The coordinates of P and Q are P(3,4) and Q(–4,–3)

Equation o34f––O00P,(xy––0y)1 = xy22––yx11 (x–x1)
or, y–0=

⸫ 3y= 4x –3–0
–4–0
Equation of OQ, y–0= (x–0)

⸫ 4y= 3x

Hence the required equations are 4x = 3y and 4y = 3x.

35. In the figure ABC is a triangle with vertices A(2,2), B(2,8) and C(6,–2). Then

Solution
(i) Find the coordinates of mid points P of AB and Q of AC

Coordinates of mid point P at AB, 2+2 , 2+8 A(2,2)
2 2 Q
= (2,5) C(6,–2)

Coordinates of mid point Q of AC, 2+6 , 2–2 P
2 2
= (4,0)

(ii) Find the equation of PQ B(2,8)

The equation of PQ
y–y1 = xy22––xy11 (x–x1)

104 Vedanta Optional Mathematics Teacher's Guide ~ 9

or, y–5= 0–5 (x–2)
4–2
–5
or, y–5= 2 (x–2)

or, 2y–10= –5x+10

⸫ 5x+2y = 20, is the required equation of PQ.

(iii) To show that PQ is parallel to BC

Slope of PQ= xy22––yx11 = 0–5 = –5
4–2 2

Slope of BC= xy22––yx11 = –2–8 = –10 –5
6–2 4= 2

Slope of PQ= Slope of BC

∴ PQ is parallel to BC.

(iv) Find the equation of perpendicular bisector of PQ

Mid point of PQ, x1 + x2 , y1 + y2
2 2

= 2+4 , 5+0
2 2

= 3, 5 , slope of PQ(m1) = – 52
2
2
slope of perpendicular bisector (m2) = 5

Now, equation of perpendicular bisector is y–y1 = m(x–x1), where m2 = 2
5 2 (x – 3) 5
y – 2 = 5

or, 4x – 10y + 13 = 0

⸫ 4x – 10y + 13 = 0 is the equation of perpendicular bisector of PQ.

36. The length of perpendicular drawn from the point (k,3) on the line 3x+4y+5= 0 is 4.
Find the value of k.

Solution
Given equation of line

3x+4y+5= 0

Point (x1,y1) = (k,3), A= 3, B= 4, C= 5
The length of perpendicular= 4

Now, d= Ax1+By1+C
A2+B2

or, 4= ± 3.k+4.3+5
32+42
3k+12+5
or, 4= ± 5

or, 20= ±(3k+17)

Taking positive sign,

Vedanta Optional Mathematics Teacher's Guide ~ 9 105

20= 3k+17

or, k= 1

Taking negative sign,

20= –(3k+17)

or, 3k= –37

or, k= –37
⸫ k = 1, –337

3

37. If the length of perpendicular form of point (1,1) to the line ax–by+c= 0 is 1, then prove that
11 c 1
c + a = 2ab + b

Solution
perpendicular distance(d)= 1

point (x1,y1) = (1,1)
Equation of line is ax–by+c= 0

∴ A = a, B = –b, C = c

Now, d= Ax1+By1+C
A2+B2

or, 1= a.1–b.1+c
a2+(–b)2

or, 1²= (a–b+c)²
a²+b²

or, a²+b²= (a–b)²+2(a–b).c+c²

or, a²+b²= a²–2ab+b²+2ca–2bc+c²

or, 2ab+2bc= c²+2ca

Dividing both sides by 2abc

or, 2ab 2bc c² 2ca
2abc + 2abc = 2abc + 1abc

11 c 1
⸫ c + a = 2ab + b
proved

38. Transform the equation xy =1 into normal form and show that 1 + 1 = 1 where
a+b a² b² p²
p is the perpendicular distance of the line from the origin.

Solution x y
a b
The line + =1

or, x + y –1= 0
a b
comparing with Ax+By+C= 0

A= 1 , B=b1 , C= –1
a

106 Vedanta Optional Mathematics Teacher's Guide ~ 9

The point (x1, y1) = (0,0)
The perpendicular distance= p

Now d= Ax1+By1+C
A2+B2

1 .0+ b1 .0–1
a
or, p=
1 1
a2 + b2

or, p²= –1 2

1 + 1
a2 b2

or, 1 = 1 + 1
p² a² b²

39. If pand p' are perpendicular distance of the line from the origin upon the straight lines whose
equations are x.secθ +ycosesθ = a and xcosθ – ysinθ = a cos²θ prove that up²+(2p')² = 4a²cos²θ

Solution
The equation of first line is
xsecθ +ycosecθ = a

or, xsecθ +cosecθ –a= 0
Distance from origin(0,0)

P= secθ×0+cosecθ×0–a
sec2θ+cosec2θ
a²
or, p²= sec² + cosec²

= a2

1 + 1
cos²θ sin²θ

or, p²= a2
sin²θ+cos²θ
cos²θ. sin²θ

or, p²= a²cos²θ .sin²θ

The equation of second line is

x.cosθ – ysinθ = acos²θ

or, x.cosθ –ysinθ –acos²θ = 0

The perpendicular distance from the origin

cosθ×0–sinθ×0–a cos2θ
p'= cos2θ + (–sinθ)2

–acos2θ

p'= 1

Vedanta Optional Mathematics Teacher's Guide ~ 9 107

p'= –acos²θ
Now, 4p²+(2p')²= 4a²cos²θ .sin²θ +(2(–a cos²θ ))²
= 4a²cos²θ .sin²θ +4a²cosθ
= 4a²cos²θ (sinθ +cos²θ)
= 4a².cos²θ .1
∴ 4p²+(2p')²= 4a²cos²θ proved

40. (a) If the points (a,0), (0,b) and (x,y) are collinear prove that x + y = 1
a b

Solution
Area of triangle formed by the vertices (a,0), (0,b) and (x,y) is given by

∆= 1 a 0x a
2 0 by 0
1
∆= 2 (ab–0+0–bx+0–ay)

Since the points are collinear, ∆= 0,

∴ 0=12 (ab–bx–ay)
or, 0= ab–bx–ay

or, bx+ay= ab

or,xbabx + ay
⸫ a+ y ab = 1
b = 1 Proved

(b) If the points (h,0), (0,k) and (4,4) lie on the same straight line, then show that + =

Solution h04 h
∆= 12 0 k4 0
1
or, 0= 2 [hk–0+0–4k+0–4h] [ ∴ ∆= 0]

or, 0= hk–4k–4h

or, 4k+4h= hk

or, 4x 4y hk (dividing both side by 4hk)
4hk + 4hk = 4hk
1 1k = 1
⸫ h + 4 proved

41. (a) The coordinates of three points are A(–6,3), B(–3,5) and C(4,–2) respectively. If P(x,y)

be any point then prove that ∆PBC = x+y–2
∆ABC 5
Solution :

Here, area of ∆PBC= 21 x –3 4 x
y 5 –2 y

= 1
2 [5x+3y+6–20+4y+2x]

= 1
2 [7x+y–14]

108 Vedanta Optional Mathematics Teacher's Guide ~ 9

Area of ∆ABC= 12 –6 –3 4 –6
3 5 –2 3
1
= 2 [–30+9+6–20+12–12]

= 1 [–35]
2
35
= 2

∴ ∆PBC = 1 (7x+y–4) = 7(x+y–2
∆ABC 2 35/2 35

∴ ∆PBC = x+y–2 Proved
∆ABC 5

(b) If (x,y) be any point in the line passing through the points p ,0 , 0, –p
cosα sinα

then prove that xcosα +ysinα = p

Solution

Here Area of triangle with vertices (x,y), p , 0 and 0, –p ) is given by
cosα sinα
∆= 12 x p/cosα 0 x
y 0 –p/sinα y

They lie on the same line, ∴ ∆= 0

or,

or, 0= 1 x p/cosα 0 x
2 y 0 –p/sinα y

or, 0 = 1 0– py – p2 –0+0+ px
2 cosα cosα.sinα sinα

or, 0= py p² + px
cos α – cosα .sinα sinα
p² py px
or, cosα.sinα = cosα + sinα

or, p² pysinα + px cosα
sinα.cosα = sinα.cosα

or, p²= p(ysinα+xsinα)
⸫ xcosα+ysinα= p Proved

43.(a) If A(k,–2), B(4,0), C(6,–3) and D(5,–5) are the four points and ∆ABC= ∆ACD in area
find the value of k.

Solution of ∆ABC== 2121 k 4 6k
Area –2 0 –3 –2
|0+8–12–0–12+3k|


= 1
2 |3k–16|
1
= 2 |3k–16|

Vedanta Optional Mathematics Teacher's Guide ~ 9 109

Area of ∆ACD= 21 k 6 5k
–2 –3 –5 –2

= 1
1 2 |–3k+12–30+15–10+5k|

= 2 |2k–13|

By question,

Area of ∆ABC= Area of ∆ACD

or, 1 1
2 |3k–16| = 2 |2k–13|
or, 3k–16= ±(2k–13)

Taking positive sign, we get

3k–16= 2k–13

or, k= 3

Taking negative sign, we get

3k–16= –2k+13

or, 5k= 29 ∴ k= 29
5
29
∴ k= 3, 5

(b) A(6,3), B(–3,5), C(4,–2) and D(a,3a) are four points. If ∆DBC = 1 , then find the coordinates
of point D. ∆ABC 2

Solution 6 –3 4 6
Area of ∆ABC= 21 3 5 –2 3
1
= 2 |30+9+6–20+12+12|

= 1
2 |49|
A rea o=f ∆=12DB|4C259a=+912a+63a–a20–5+31–242+13a2aa+2a|
1
= 2 |28a–4|
= ± (28a–14

2
= ±(14a–7)

By question,

∆DBC = 1
∆ABC 2

or, ±(14a–7) = 1
49/2 2
14a–7 1
Taking positive, 49 ×2 = 2

or, 2a–1 ×2 = 1
7 2

110 Vedanta Optional Mathematics Teacher's Guide ~ 9

or, 8a–4= 7

or, 8a= 11

or, a= 11
8
–(14a–7) 1
Taking negative. 49/2 ×2 = 2

or, –2(2a–7) ×2 = 1
7 2
or, –8a+28= 7

or, –8a= –21

11 ⸫ a= 21
21 8

∴ a= 8, 8

∴ (a,3a)= 181, 33 , or 281, 63
8 8

(c) A and B are two points with coordinates (3,4) and (5,–2) respectively. Find a point P such
that PA= PB and ∆PAB= 10 sq. units.

Solution
Let the coordinates of P be (x,y)

coordinate of A= (3,4)

coordinate of B= (5,–2)

By question,

PA= PB

PA²= PB²

or, (x–3)²+(y–4)²= (x–5)²+ (y+2)²

or, x²–6x+9+y²–8y+16= x²–10x+25+y²+4y+4

or, 4x–12y= 4

or, x–3y= 1

or, x= 1+3y .................(i)

∴ The coordinates of P are (x,y)= (1+3y,y)

Area of ∆PAB= 12 1+3y 35 1+3y
y 4 –2 y
1
or, 10= 2 |4(1+3y)–3y–6–20+5y+2(1+3)|

or, 20= |4+12y–3y–26+5+2+6y|

or, 20= |20y–20|

⸫ 20= ±(20y–20)

Taking positive sign, 20= 20y–20

or, y= 2

Taking negative sign, 20= –(20y–20)

or, 20= –20y+20

or, y= 0

The point is (x,y)= (1+3y)= (7,2) when y= 2

Vedanta Optional Mathematics Teacher's Guide ~ 9 111

or, (1+3y,)= (1,0) when y= 0
⸫ Required point P is (7,2) or (1,0).

(44) In ∆ABC, with vertices A(4,5), B(–2,1) and C(0,3), E and F are the mid points of AB and
AC respectively. Then

(a) Find the coordinates of E and F

(b) Find the area of quadrilateral BCFE A(4,5)
Solution

For mid point of E of AB,

(x,y)= 42–2, 5+1 EF
2
= (2,3)
For mid point of F of AC,

(x,y)= 4+2 0, 5+3 B(–2,1) C(10,3)
2
= (2,4)

∴The coordinates of E= (2,3)
The coordinates of F= (2,4)

Area of quadrilateral BCFE

= 1 –2 0 2 2
2 13 4 3
1
= 2 |–6–0+0–6+6–8|

= 1 | –14|
2
=7

45. If P,Q, and R are the mid point of the sides BC, CA and AB of ∆ABC whose vertices
∆PQR 1
A(1,–4), B(5,6) and C(–3,3). Show that ∆ABC = 4

Solution

The coordinates of P= 5–3 6+3 A(1,–4)
2, 2 RQ

= (1, 9 )
2
3–4
The coordinates of Q= –3+1 ,
2 2
= (–1, –21)
–4+6
The coordinates of R= 1+5 ,
2 2
= (3,1) B(5,6) P C(–3,3)

Area of ∆ABC= 1 1 5 –3 1
2 –4 6 3 –4
1
= 2 |6+20+15+18+12–3|

112 Vedanta Optional Mathematics Teacher's Guide ~ 9

= 1 |68|
2

= 34 sq.unit

Area of ∆PQR= 12 1 –1 3 1
9/2 6 1 9/2

= 1 –1 + 9 –1 + 3 + 27 –1
2 2 2 2 2

= 1 –1 +9 –2 +3 +27 -2
2 2

= 1 |324|
2
17
= 2

Now ∆PQR = 17/2 = 1 proved.
∆ABC 34 4

Questions for practice

1. Find the equation of a straight line passing through the point (2, 3) and making equal
intercepts on the axes.

2. Find the equation of a straight line passing through the point (0, 7) and making an angle
of 135° with X-axis in positive direction.

3. Find the equation of the angle bisector between the axes.
4. Find the equation of straight line which passes through the point (4, 8) and whose

x-intercept on X-axis is three times that its intercept on Y-axis.
5. Find the equation of a straight line passes through the point (2, 3) and the portion of

the line intercepted between the axes is bisected at the point, find the lenght of the
intercepted portion.
6. Find the equation of the straight line passes through the point of intersection of
3x + y = 6 and 2x – y = 4 and (3, 4).
7. Find the equation of a straight line making an angle of 60° with the X-axis and intersecting
the Y-axis at a distance of 5 units.
8. If p and p1 are the lengths of the perpendiculars from the points (± a2 – b2, 0) to the line

x cosθ + y sinθ, prove that: p × p1 = b2.
a b

9. If A, B, C and D are the four points with coordinates (6, 3), (–3, 5), (4, –2) an d(a, 3a)

respectively. If ∆DBC = 41, find the coordinates of D.
∆ABC

10. P(0, 1), Q(5, 1), R(7, 5) and S(2, 5) are the vertices of a parallelogram. Find the following:
i) Equations of the diagonals of the parallelogram.
ii) Area of the parallelogram.

Vedanta Optional Mathematics Teacher's Guide ~ 9 113

UNIT

eight Trigonometry

Measurement of Angles Estimated Teaching Hours : 7

1. Objectives

SN Level Objectives

To tell different systems of measurement of angles

(sexagesimal, centesimal and radian)
i Knowledge (K) Define 1c
lc
To tell formula θ = r and meaning of the symbols used.

To convert measure of angles.
ii Understanding (U) Sexagesimal into centesimal and vice.

Radian into sexagesimal and centesimal and vice-versa.

To solve verbal problems of measurement of angles.

iii Application (A) Apply formula θ =  l c to solve problems.
r 

To derive the following:

iv Higher Ability (HA) Radian is a constant angle, θ =  l c
 r 

2. Teaching Materials
Relation chart of measurement of angles.
3. Teaching Learning Strategies:
Review concept of an angle.
Discuss the different measure of angles - sexagesimal, contesimal and radian and

relation between them.
Illustrate examples - conversion of angles degree into grade i.e. sexagesimal into

centesimal grade into degree i.e. centesimal into sexagesimal.
Solve some questions from exercise 8.1 as examples and let the students do some

questions and the teacher supervise them.
Discuss about the circular measure of angles.
Illustrate with examples conversion of sexagesimal and centesimal into radian measure

and vice versa.
Derive the formula,

θ =  l c
r 

Give clear meaning of θ, l and r and their units.

114 Vedanta Optional Mathematics Teacher's Guide ~ 9

The teacher gives some examples solution of problem by using the above formula.
The teacher uses some chart to derive the formula.
Let the students do some problems in the class and the teacher provides necessary guidance.
Notes:
1) Sexagesimal or British measure
60 seconds (60") = 1 minute (1')
60 minutes (60') = 1 degree (1°)
90 degree (90°) = 1 right angle.
2) Centesimal or French measure
100 seconds (100") = 1 minute (1')
100 minutes (100') = 1 grade (1g)
100 grades (100g) = 1 right angle.
3) Relation between degree and grade

90° = 100g ⇒ 1° = 10g and 1g =  9° 
9 10
D G
Also, 9 = 10

Where, D = no. of degrees and G = no. of grades.
4) Relation between degree, grade and radian
180° = 200g = πc

or, 1° = 1π8c0, 1g = πc
200

or, 1c =  180° , 1c=  200 g
 π  π 

5) Relation between arc length, central angle and radius

θ =  l c ⇒ l = θr ⇒ r = l
 r θ

Where l = arc length, r = radius and θ = angle at the centre of circle

Here, l and r must be in same unit.

θ must be in radian.

6) D = G = 2c
90 100 π

Where, D = degree, G = grade and C = radian
7) Interior angle of a regular polygon

θ = n – 2 × 180°
n

where n = number of sides.
Exterior angle of a regular polygon

α = 360°
n

Vedanta Optional Mathematics Teacher's Guide ~ 9 115

Some solved problems

1. Reduce 20g 10g 20" into degree, minutes and seconds.

Solution
Here, 20g 10g 20"

= 20g +  10 g +  20 g
 100  10000
1g = 190° 
= 20.1020g 
9°
= 20.1020 × 10

= 10° 5' 30"

2. Find the ratio of 81° and 30g.

Solution

Here, 81°

and 30g = 30 × 9° = 27°
10

Now, ratio of 81° and 30g = 81°
30g

= 81° = 3:1
27°

3. Two angles of a triangle are 40g and 120g. Find the remaining angle in degrees.

Solution

Here, 40g = 40 × 9° = 36°
10

and 120g = 120 × 9° = 108°
10

Let x° be the remaining angle.
Then, sum of angles of a triangle = 180°
i.e. 36° + 108° + x° = 180°
or, x° = 180° – 144°
 x° = 36°
4. The angles of a triangle are in the ratio of 2:3:4. Find the all the angles in degrees and grades.

Solution

Let, 2k, 3k and 4k be the angles of the given triangle

Then, sum of angles of a triangle = 180°

2k + 3k + 4k = 180°

or, 9k = 180°

 k = 20°

In degree In grade

2k = 2 × 20° = 40° 40° = 40 × 10g = 44.44g
9

116 Vedanta Optional Mathematics Teacher's Guide ~ 9

3k = 3 × 20° = 60° 60° = 60 × 10g = 66.67g
9

4k = 3 × 20° = 80° 80° = 80 × 10g = 88.89g
9

5. a) Through what angle does the minute hand of a clock turn in 25 minutes.

Solution
Here, In a clock,
The minute hand turn through 360° in 60 minutes

The minute hand turn through 36600° = 6˚ in 1 minute

6° × 25 in 25 minutes
= 150°

b) Through what angle does the hour hand of a clock turn in 4 hours ?

Solution
Here, In a clock
The hour hand turns through 360° in 12 hours

The hour hand turns through 31620° = 30˚ in 1 minute

30° × 4 in 4 minutes
= 120°

6. Find the ratio of 20° and 23πc.
Solution

Here, 2πc = 2π × 180° = 120°
3 3 π

Now, ratio of 20° and 23πc = 20° : 2πc
3

= 20° : 120°

= 1:6

7. The difference of two acute angles of right angled triangle of right angled triangle is π6c.
Find all the angles in radian.

Solution
Let, two acute angles be xc and yc.

Then, xc – yc = πc ... ... ... (i)
6
πc
and xc + yc = 90° = 90 × 180

 xc + yc = πc ... ... ... (ii)
2

Solving (i) and (ii), we get

Vedanta Optional Mathematics Teacher's Guide ~ 9 117

xc = πc and yc = πc
3 6
πc
Also, 90° = 2

8. Find the length of an arc of a circle of radius 21cm of the angle subtended at the centre

by the arc is 30°.  π = 22  .
 7 

Solution

Here, angle at the centre (θ) = 30° = 30 × πc = πc
6 6
π = 272, radius (r) = 21cm

arc length (l) = ?

By using formula, l = rθ

= 21 × πc
6

= 21 × 22 × 1 = 11cm
7 6

9. Three angles of a triangle are  20x g, 3x° and  πx c. Find all the angles in degrees.
9 45

Solution

Here,  20x g = 20x × 9° = 2x°
9 9 10

 πx c = πx × 180° = 4x°
45  45 π

We know that

sum of angles of a triangle = 180°

i.e. 2x° + 4x° + 3x° = 180°

or, 9x° = 180°

 x° = 20°

Required angles are 2x° = 2 × 20° = 40°

3x° = 3 × 20° = 60°

4x° = 4 × 20° = 80°

10. Find the angle between the minute hand and the hour hand in radian at four o'clock.

Solution 11 12 1
At four o'clock the minute hand is at 12. 10 2
and hour hand is at 4.
Angle between them = 20 minute 93
60 minute = 360°

1 minute = 360° = 6° 84
60 76 5

20 minute = 6° × 20 = 120°

118 Vedanta Optional Mathematics Teacher's Guide ~ 9

= 120 × πc
180

= 2πc
3

11. Find the interior and exterior angles of a regular hexagon in degrees and grades.

Solution
In a regular hexagon
number of sides (n) = 6
Interior angle is given by

θ = n – 2 × 180°
6

= 6 – 2 × 180° = 120°
6

In grade, θ = 120 × 10°
9

= 133.33g

Exterior angle is given by, α = 360°
n

= 360° = 60°
6

In grade, θ = 60 × 10 = 66.67g
9

12. a) The difference of the number of grades and the number of degree of the same angle
is 16. Find the number of degrees and grades of the angle.

Solution
Let, x be measure in grade and y be in degree.
Since number of grade is greater than number of degree of the same angle.
Then by question
x – y = 16 ... ... ... (i)
Again, xg = y°

xg = y × 10g
9

 x = 10y ... ... ... (ii)
9

put the value of x in equation (i), we get

10y – y = 16
9

or, y = 144
put the value of y in (ii), we get

Vedanta Optional Mathematics Teacher's Guide ~ 9 119

x = 10 × 144° = 160°
9

 x = 160g and y = 144°.

b) The number of degrees of a certain angle added to the number of grades of the same
angle is 152, find the angle in degrees and grades.

Solution
Let, x and y be number of degrees and grades of the same angle respectively.
Then, x° + yg = 152 ... ... ... (i)

we have, x° = yg =y× 9°
10

put the value of x° in (i), we have

91y0g + yg = 152

or, 19yg = 152 × 10

or, yg = 1520 = 80
19

 y = 80g

and x = 9y = 9 × 80 = 72°
10 10

 x = 72° and y = 80g.

13. In a triangle, the first angle is greater than the second by 18° and less
than the third by 10g. Express all the angles in degrees.

Solution
Let, x° be the first angle of the given triangle.
Then, second angle (y) = x° – 18°

and 10g = 10 × 9° = 9°
10

third angle (z) = x° + 10g = x° + 9°

Then, sum of the angles of a triangle = 180°

x° + y° + z = 180°

or, x° + x° – 18° + x° + 9° = 180°

or, 3x° = 189°

 x° = 63°

In degree In grade

First angle (x°) = 63° 63° = 63 × 10g = 70g
Second angle (y°) = 63° – 18° = 45° 9
Third angle (z) = 63° + 9° = 72°
45° = 45 × 10g = 50g
9

72° = 72 × 10g = 80g
9

120 Vedanta Optional Mathematics Teacher's Guide ~ 9

14. a) If D and G are the number of degrees and number of grades of the same angle, then

prove that: G = D + D
9

Solution

Here, G and D are the number of grades and degrees of the same angle

Gg = D°

or, Gg = 10 Dg
9

 G = D + D proved.
9

b) If S1 and S2 are the number of sexagesimal and centesimal seconds of any angle, then

prove that: S1 = 2S520.
81

Solution

Here, S1 (sexagecimal second) = S2 (centesimal second)

60 S1 60 = 100 S2 100
× ×

or, S91 = S2 × 9°
25 10

 S1 = S2 proved.
81 250

15. The number of sides of two regular polygons are in the ratio of 5:4. If the difference of
their interior angles is 9°, find the number of sides of each polygon.

Solution

Let, n1 and n2 be the number of sides of the first and second regular polygons respectively.

then, nn21 = 45 ⇒ n1 = 5 n2 ... ... ... (i)
4

Again, interior angle of a regular polygon = n – 2 × 180°
θ1 – θ2 = 9° n

or, n1n–1 2 × 180° – n2n–2 2 × 180° = 9°

or, nn11 – 2 – n2 + 2 × 180° = 9°
n1 n2 n2

or, 1 – 2 – 1 + 2 = 9°
n1 n2 180°

or, n22 – 2 = 1 ... ... ... (ii)
n1 20

put the value of n1 from (i) in (ii), we get

Vedanta Optional Mathematics Teacher's Guide ~ 9 121

2 – 2 = 1
n2 20
5 n2
4

or, 105n–28 = 1
20

or, (10 – 8) . 4 = n2
 n2 = 8
put the value of n2 in (i), we get

 n1 = 5 × 8 = 10
4

 n1 = 10 and n2 = 8.

16. One regular polygon has twice as many sides as another, if the ratio of interior angles
of the first to that of the second is 5:4. Find the number of sides of each polygon.

Solution
Hints: n1 = 2n2 ... ... ... (i)

θ1 = 45 ⇒ n1n–1 2 × 180° = 5 ... ... ... (ii)
θ2 4
n2 – 2 × 180°
n2
 θ1 = n1n–1 2 × 180°
 θ2 = n2n–2 2 × 180°
solving (i) and (ii), we get

n1 = 12, n2 = 6

17. A goat is tethered to a stake by a rope 4.5m long. If the goat moves along the circumference

of a circle always keeping the rope tight. Find how far it will have gone when the rope

has traced out an angle 150°. π = 22 
7 
Solution
A
Angle at centre (θ) = 150°

= 150 × πc = 5πc
radius 4.5m 180 6
(r) = 4.5m

arc length (l) = ?

By using formula O 150˚ l
l = θr

= 5πc × 4.5
6

= 5 × 22 × 4.5 = 11.79m
6×7

 The goat has moved through 11.79m. B

122 Vedanta Optional Mathematics Teacher's Guide ~ 9

18. The number of degree of an angle of a triangle exceeds that of the second by 20° and

the circular measure of the third exceeds that of the second by  π c. Find all the angles
18
in degrees.

Solution

Let, ∆ABC be given triangle A
BC
Let, B = y

A = y + 20°

C = y + 10°

πc = π × 180 = 10°
18 18 π

Since sum of angles of a triangle = 180°

we have, y + y + 20° + y + 10° = 180°

or, 3y = 150°

 y = 50°

 A = y + 20° = 50° + 20° = 70°

B = y = 50°

C = y + 10° = 50° + 10° = 60°

19. Divide 80° into two parts such that the number of degrees in the first to the number of
radian to the second is 300:π. Find the angles in degrees.

Solution

Let, x and y be the required two angles.

Then, let x = 300k°

and y = πkc = πk × 180° = 180k°
π

Sum of two angles is 80°

i.e. x + y = 80°

or, 300k° + 180k° = 80°

or, 480k° = 80°

 k = 1
6

 x = 300 × 1 = 50° and y = 180 × 1 = 30°
6 6

20. If D, G and C are the number of degree, grade and radians of an angle, prove that:

G = C = D
200 π 180

Solution

We have,

Gg = Cc = D°

or, G × 9° = C × 180° = D°
10 π

Vedanta Optional Mathematics Teacher's Guide ~ 9 123

Dividing each side by 180, we get

G = C = D proved.
200 π 180

21. A bus is traveling a circular tturarcnkedofin53 km radius at 44km per hour speed. Find the
angle in degree through which it a minute.

Solution

Here, radius (r) = 5 km = 5000 m
3 3

speed = 44 km/hr

= 44 × 1000m
60 min

= 2200 m/s
3

i.e. the bus travels 2200 metre in 1 minute. it will be an arc length on the circular track for
3

angle θ (suppose)

 l = 2200 m
3

2200

θ = rl = 3
5000

3
11c
= 25

= 11 × 180°
25 π

= 11 × 180 × 7° = 25° 12'
25 22

22. A wheel turns 30 revolutions in one minute. Through how many (i) degrees (ii) radians
does it turn in one second ?

Solution
Here, In 1 minute the wheel turns 30 revolutions
i.e. in 60 seconds the wheel turns 30 revolutions

in 1 second = 1 turn
2

i) 1 revolution = 360°

 1 revolution = 1 × 360°
2 2

= 180°

ii) Also, 180° = πc.

124 Vedanta Optional Mathematics Teacher's Guide ~ 9

23. Find the degrees the angle between the hour hand and minute hand of a clock.
a) Quarter past two

Solution

At quarter past two, the hour hand and the minute hand of the 12

clock are as shown in the figure. P
O Q3
Angle between them is given by (taking acute angle)

θ = POQ

The minute hand turns through = 6° × 15 9

(60 minutes = 360°) = 90°

The hour hand turns through = 30° × 2 + 1 6
4

(12 hour = 360°) = 30° × 9
4

= 67.5°

 The required angle is (90° – 67.5°) = 22.5°.

b) Quarter to six.

Solution 9Q 12 3
The positions of the hour hand and the minute hands are shown O
P
in the figure. 6
we know that the minute hand turns through 360° in 60

minutes and the hour hand turns through 360° in 12 hour.
In the figure,
The minute hand is turned through 45 × 6° = 270°

The hour hand is turned through 30° × 5 + 34 = 172.5°

 The angle between the hour hand and minute hand at quarter to six is given by

POQ = (270° – 172.5°)

= 97.5° = 97° 30'

c) at 6:30 PM

Solution 12
The positions of the hour hand and the minute hand are as shown
O
in the figure. Q
The minute hand turns through 6° × 30 = 180°
P
The hour hand turns through 30° ×  6 + 1 = 195° 9 6 3
 2

Hence, the required angle is given by
POQ = 195° – 180°
= 15°

Vedanta Optional Mathematics Teacher's Guide ~ 9 125

Questions for practice

1. Two angles of a triangle are 60° and 70g, find the third angle in circular measure.
2. One angle of a triangle is 40g and the remaining angles are in the ratio of 5:7, find their

measure in degrees.
3. Find the exterior and interior angles in grade of the following regular polygons
a) Dodecagon
b) Quin decagon
4. The number of degrees in an angle of a triangle to the number of grade in second is

to the number of radian in third is in the ratio of 900:1200, 7π, find all the angles in
degrees.
5. An angle of 105° is divided into two pars such that the ratio of the number of grades in
the first part to the number of degrees in the second part is 5:6, find he angles in circular
measure.
6. A goat is tied to a pole by a rope of 24 feet long. It moves keeping the rope tight. How

far will it reach when the rope traced out an angle of 3πc ?
8

7. Taking the radius of the earth as 6371 km; find the distance between two places whose
latitudes are differed by 63°.

8. A car travelling in a circular track at a speed of 90 km/hr. What angle does it subtands
in 5 seconds if the radius of track is 2.4 km?

9. The moon is 360000 km far from the earth. The diameter of the moon subtends an angle
of 31' (minute) at an observer's eye, find the diameter of the moon.

10. The length of a pendulum is 7cm. If the pendulum swings through 12° 30' to each side
of its mean position, find the length of arc through which the tip of pendulum passes.

126 Vedanta Optional Mathematics Teacher's Guide ~ 9

UNIT

nine Trigonometric Ratios &
Conversion of t-ratios

Symbol used t - trigonometric Estimated Teaching Hours : 8

1. Objectives

SN Level Objectives

To define t - ratios.

i Knowledge (K) To define t - identities .
To do factorization of simple t - expressions

To find product of simple t - factors.

ii Understanding (U) To prove simple t - identities.
To convert t - ratios into different forms.

iii Application (A) To prove t - identies.
To find value of given t - expressions using given t - ratios.

iv Higher Ability (HA) To prove harder t - identies.

2. Teaching Materials
Trigonometric ratio table.

Table of conversion of trigonometric ratios.

Formula table of algebra for factorization.

3. Teaching Learning Strategies:
Review the concept of t - ratios.
Derive trigonometric identities from Pythagoras theorem.
Discuss about reciprocal relations among t - ratios.
Illustrate the process of factorization of trigonometric expressions with comparing algebraic
expressions.
Review the following formula from algebra
(a + b)2, (a – b)2, a2 – b2, a3 + b3, a3 – b3, (a + b)3, (a – b)3
Solve some problems from exercise 9.1
Discuss about trigonometric identities and how to prove trigonometric identities.
Solve some problems from exercise 9.2
Convert given trigonometric ratios into the remaining trigonometric ratios with at least two
examples.
Solve some problems from exercise 9.3 with illustration.

Vedanta Optional Mathematics Teacher's Guide ~ 9 127

Notes:
1) a2 – b2 = (a + b) (a – b)

2) a3 – b3 = (a – b) (a2 + ab + b2)

3) a3 + b3 = (a + b) (a2 – ab + b2)

4) i) sin2θ + cos2θ = 1 ⇒ sinθ = 1 – cos2θ
ii) sec2θ – tan2θ = 1
iii) cosec2θ – cot2θ = 1 ⇒ secθ = 1 + tan2θ
iv) sinθ . cosecθ = 1
⇒ cosecθ = 1 + cot2θ
v) cosθ . secθ = 1
⇒ sinθ = 1
vi) tanθ . cotθ = 1 cosecθ
1
⇒ secθ = cosθ

⇒ tanθ = 1
cotθ

Some solved problems

1. Prove that:

a) cosecθ . cotθ . 1 + tan2θ . sin3θ

Solution
LHS cosecθ . cotθ . 1 + tan2θ . sin3θ
= cosecθ . cotθ . secθ

= si1nθ . cosθ . 1 . sin3θ
sinθ cosθ

= sinθ

b) (tanθ + secθ)2 = 1 + sinθ
1 – sinθ

Solution
LHS (tanθ + secθ)2

= csoinsθθ + 1 2
cosθ 

= (sincθos+2θ1)2

= (sinθ + 1)2
1 – sin2θ

= (1 + sinθ) (1 + sinθ) = 1 + sinθ = RHS proved.
(1 – sinθ) (1 + sinθ) 1 – sinθ

128 Vedanta Optional Mathematics Teacher's Guide ~ 9

c) 1 cotθ = cosecθ – cotθ
cosecθ +
Solution

LHS 1
cosecθ + cotθ

= 1 cotθ × cosecθ – cotθ
cosecθ + cosecθ – cotθ

= cosecθ1– cotθ
= cosecθ – cotθ = RHS proved.

d) 1 + cosθ = cosecθ + cotθ
1 – cosθ

Solution

LHS 1 + cosθ
1 – cosθ

= 1+ cosθ × 1 + cosθ
1– cosθ 1 + cosθ

= (1 + cosθ)2
1 – cos2θ

= 1 +sincθosθ

= 1 + cosθ
sinθ sinθ

= cosecθ + cotθ = RHS proved.

e) 1 sinθ + 1 sinθ = 2 cosecθ
– cosθ + cosθ
Solution

LHS 1 sinθ + 1 sinθ
– cosθ + cosθ

= sinθ  1 – 1 + 1 1 
 cosθ + cosθ 

= sinθ  1+ cosθ + 1 – cosθ 
 (1 – cosθ) (1 + cosθ) 

= sinθ . 1 – 2
cos2θ

= sinθ . 2
sin2θ
= si2nθ

= 2 cosecθ = RHS proved.

Vedanta Optional Mathematics Teacher's Guide ~ 9 129

2. a) If cosθ = x x 1, find the value of cotθ.
+
Solution

Here, cosθ = x x 1
+

cosθ = cosθ
sinθ

= cosθ
1 – cos2θ

= x
x+1

1 – (x x2
+ 1)2

= x × x+1

x+1 x2 + 2x + 1 – x2

= x
2x + 1

b) If sinθ + cosecθ = 3, find the value of sin2θ +cosec2θ

Solution
Here, sinθ + cosecθ = 3
squaring on both sides,
sin2θ + 2 sinθ . cosecθ + cosec2θ = 9
or, sin2θ + cosec2θ + 2.1 = 9
 sin2θ + cosec2θ = 7.

3. Express tanθ in terms of remaining trigonometric ratios.

Solution

Here, i) tanθ 1 – cos2θ
cosθ
ii) tanθ = sinθ =
cosθ

iii) tanθ = sinθ
1 – sin2θ

iv) tanθ = 1
cotθ

v) tanθ = 1 = 1
cotθ 1 – cosec2θ

vi) tanθ = sec2θ – 1

3. Prove the following identities
a) (sin3θ + cos3θ) = (sinθ + cosθ) (1 – sinθ cosθ)

130 Vedanta Optional Mathematics Teacher's Guide ~ 9

Solution
LHS sin3θ + cos3θ
= (sinθ + cosθ) (sin2θ – sinθ . cosθ + cos2θ)
= (sinθ + cosθ) (sin2θ + cos2θ – sinθ . cosθ)
= (sinθ + cosθ) (1 – sinθ cosθ)
 LHS = RHS proved

b) sin6θ – cos6θ = (2 cos2θ – 1) (1 – sin2θ cos2θ)

Solution
LHS sin6θ – cos6θ
= (sin2θ)3 – (cos2θ)3
= (sin2θ – cos2θ) (sin4θ + sin2θ . cos2θ + cos4θ)
= (1 – cos2θ – cos2θ) (sin4θ + cos4θ + sin2θ . cos2θ)
= (1 – 2 cos2θ) ((sin2θ + cos2θ)2 – 2sin2θ . cos2θ + sin2θ . cos2θ)
= (2 cos2θ – 1) (1 – sin2θ cos2θ)
 LHS = RHS proved

c) sin8θ – cos8θ = (sin2θ – cos2θ) (1 – 2 sin2θ cos2θ)

Solution
LHS sin8θ – cos8θ
= (sin4θ)3 – (cos4θ)3
= (sin4θ – cos4θ) (sin4θ + cos4θ)
= [(sin2θ)2 – (cos2θ)2] [(sin2θ)2 + (cos2θ)2]
= (sin2θ + cos2θ) (sin2θ – cos2θ) [(sin2θ + cos2θ)2 2 – sin2θ . cos2θ]
= 1 . (sin2θ – cos2θ) (12 – 2 sin2θ . cos2θ)
= (sin2θ – cos2θ) (1 – 2 sin2θ cos2θ)
 LHS = RHS proved

d) cosec6θ – cot6θ = 1 + 3 cot2θ . cosec2θ

Solution
LHS cosec6θ – cot6θ
= (cosec2θ)3 – (cot2θ)3
= (cosec2θ – cot2θ) (cosec4θ + cosec2θ . cot2θ + cot4θ)
= 1 . [(cosec2θ)2 + (cot2θ)2 + cosec2θ . cot2θ]
= (cosec2θ – cot2θ)2 + 2 cosec2θ . cot2θ + cosec2θ . cot2θ
= 12 + 3 cosec2θ . cot2θ
= 1 + 3 cot2θ . cosec2θ
 LHS = RHS proved

e) (sinθ + secθ)2 + (cosecθ + cosθ)2= (1 + secθ . cosecθ)2

Solution
LHS (sinθ + secθ)2 + (cosecθ + cosθ)2
= sin2θ + 2 sinθ . secθ + sec2θ + cosec2θ + 2 cosecθ . cosθ + cos2θ

Vedanta Optional Mathematics Teacher's Guide ~ 9 131

= sin2θ + cos2θ + 2 (sinθ . secθ + cosecθ . cosθ) + sec2θ + cosec2θ

= 1 + 2  sinθ + cosθ + 1 + 1
cosθ sinθ  cos2θ sin2θ

= 1 + 2 . (sin2θ + cos2θ) + sin2θ + cos2θ
cosθ . sinθ cos2θ . sin2θ

= 1 + 2 . 1 + 1

cosθ . sinθ cos2θ . sin2θ

= 1 + 2 secθ . cosecθ + sec2θ . cosec2θ
= (1 + secθ . cosecθ)2
 LHS = RHS proved

f) (x sinθ – y cosθ)2 + (x cosθ + y sinθ)2 = x2 + y2

Solution
LHS (x sinθ – y cosθ)2 + (x cosθ + y sinθ)2
= x2 sin2θ – 2x sinθ . y cosθ + y2 cos2θ + x2 cos2θ + 2x cosθ . y sinθ + y2 sin2θ
= x2 sin2θ + x2 cos2θ + y2 cos2θ + y2 sin2θ
= x2(sin2θ + cos2θ) + y2(cos2θ + sin2θ)
= x2 + y2
 LHS = RHS proved

g) (sinA + cosecA)2 + (cosA + secA)2 = tan2A + cot2A + 7

Solution
LHS (sinA + cosecA)2 + (cosA + secA)2
= sin2A + 2 sinA . cosec A + cosec2A + cos2A + 2 cosA . cosecA + sec2A
= sin2A + cos2A + 2 . 1 + 2 . 1 + cosec2A + sec2A
= 1 + 2 + 2 + 1 + cot2A + 1 + tan2A
= tan2A + cot2A + 7
 LHS = RHS proved

h) (1 + tanA)2 + (1 – tanA)2 = 2 sec2A

Solution
LHS (1 + tanA)2 + (1 – tanA)2
= 1 + 2 tanA + tan2A + 1 – 2 tanA + tan2A
= 2 + 2 tan2A
= 2(1 + tan2A)
= 2 sec2A
 LHS = RHS proved

i) (1 – cotθ)2 + (1 + cotθ)2 = 2 cosec2θ

Solution
LHS (1 – cotθ)2 + (1 + cotθ)2
= 1 – 2 cotθ + cot2θ + 1 + 2 cotθ + cot2θ

132 Vedanta Optional Mathematics Teacher's Guide ~ 9

= 1 + 1 + cot2θ + cot2θ
= 2 + 2 cot2θ
= 2(1 + cot2θ)
= 2 cosec2θ
 LHS = RHS proved

5. Prove the following identities.

a) tanq 1 – 1 sinq = 2 cotq.
secq – + cosq
Solution

LHS tanq 1 – 1 sinq
secq – + cosq

sinθ

= co1csoθsθ– – 1 sinq
+ cosq
1

sinθ

= cosθ – 1 sinq
1 – cosθ + cosq

cosθ

= sinq + sinq . cosq – sinq + sinq . cosθ
(1 – cosθ) (1 + cosq)

= 2 1sin–θco. sc2oθsθ
= 2 sinsiθn.2θcosθ
= 2 cotq.
 LHS = RHS proved

b) 1 cotq – 1 = 1 – 1 cotq
cosecq – sinq sinq cosecq +
Solution

LHS 1 cotq – 1
cosecq – sinq

= cosecq1– cotq + 1 cotq – 1 cotq – 1
cosecq + cosecq + sinq

= c(coosseeccqq+– cotq + cosecq – cotq – 1 – 1 cotq
cotq) (cosecq – cotq) sinq cosecq +

= cos2ecc2oqs–eccθot2q – 1 – 1 cotq
sinq cosecq +

= 2 . 1 – 1 – 1 cotq
sinq sinq cosecq +

Vedanta Optional Mathematics Teacher's Guide ~ 9 133

= si1nq – 1 cotq
cosecq +

 LHS = RHS proved

c) cosq + 1 sinq = sinq + cosq
1 – tanq – cotq
Solution

LHS 1 cosq + 1 sinq
– tanq – cotq

= cosq + sinq
sinq cosq
1 – cosθ 1 – sinθ

= coscθo–s2sqinq + sin2q
sinθ – cosq

= coscθo–s2sqinq – sin2q
cosθ – sinq

= ccooss2θq – sin2q
– sinq

= (cosq – sinq) (cosq + sinq)
cosθ – sinq

= sinq + cosq

 LHS = RHS proved

d) 1 tanq + 1 cotq = secq . cosecq + 1
– cotq – tanq
Solution

LHS 1 tanq + 1 cotq
– cotq – tanq

sinθ cosθ

= cosθ + sinθ
cosq sinq
1 – sinθ 1 – cosθ

= cosθ(ssininq2q– cosq) + cos2q sinq)
sinθ(cosq –

= sin3q – cos3q
sinq . cosθ(sinq – cosq)

= (sinq – cosq) (sin2q + sinq . cosq + cos2q)
sinq . cosθ(sinq – cosq)

= 1 +sinsθin.qc.ocsoqsq

= secq . cosecq + 1
 LHS = RHS proved

134 Vedanta Optional Mathematics Teacher's Guide ~ 9

6. a) secq – tanq + 1 = secq – tanq = 1 – sinq
1 + secq + tanq cosq
Solution

LHS secq – tanq + 1
1 + secq + tanq

= secq – tanq + sec2q – tan2q [ sec2q – tan2q = 1]
1 + secq + tanq

= (secq – tanq) + (secq – tanq) (secq + tanq)
1 + secq + tanq

= (secq – tanq) 1 + secq + tanq
1 + secq + tanq

= secq – tanq = MHS
Again, secq – tanq

= co1sq – sinq
cosq

= 1 c–ossiqnq

 LHS = RHS proved

b) cosecq + cotq + 1 = cosecq + cotq = 1 + cosq
1 + cosecq – cotq sinq
Solution

LHS cosecq + cotq + 1
1 + cosecq – cotq

= cosecq + cotq + (cosec2q – cot2q)
1 + cosecq – cotq

= (cosecq + cotq) + (1 + cosecq – cotq)
1 + cosecq – cotq

= cosecq + cotq = MHS
Again, cosecq + cotq

= si1nq – cosq
sinq

= 1 +sincqosq

 LHS = RHS proved

c) seca – tana + 1 = 1 + seca + tana
seca – tana – 1 1 – seca – tana
Solution

LHS seca – tana + 1
seca – tana – 1

= seca – tana + sec2a – tan2a
seca – tana – (sec2a – tan2a)

Vedanta Optional Mathematics Teacher's Guide ~ 9 135

= ((sseeccaa – tana) + (seca – tana) (seca + tana)
– tana) – (seca – tana) (seca + tana)

= (s(seeccaa––tatannaa))[1[1+– seca + tana]
seca – tana]

= 11+– seca + tana
seca – tana

 LHS = RHS proved

d) coseca + cota –1 = 1 – coseca + cota
coseca + cota +1 1 + coseca – cota
Solution

LHS coseca + cota – 1
coseca + cota + 1

= (c(ocsoesceaca++coctoata) +) (c(ocosesce2ca2a––ccoot2tq2a) )

= ((ccoosseeccaa + cota) (1 – coseca + cota)
+ cota) (1 + coseca – cota)

= 11 –+ccoosseeccaa+– cota
cota

 LHS = RHS proved

e) 1 – sinq + cosq = 1 + cosq
sinq + cosq – 1 sinq
Solution

LHS 1 – sinq + cosq
sinq + cosq – 1

= 1sin–qsi+nqc+osqco–s1q × sinq
sinq

= sin(sqin–qs+in2cqos+q sinq . cosq
– 1) sinq

= sin(qs(i1nq++cocsoqs)q––(11)–scinoqs2q)

= (1 – cosq) (sinq – 1 + cosq)
(sinq + cosq – 1) sinq

= 1 +sincqosq

 LHS = RHS proved

f) cosa + cosb = cosa + cosb
sina + cosb sinb – cosa sina – cosb sinb + cosa
Solution

LHS cosa + cosb
sina + cosb sinb – cosa

136 Vedanta Optional Mathematics MTeaancuhaelr'~s G1u0ide ~ 9

= sinaco+sacosb + cosb – sinaco–sacosb+ cosa
sinb – cosa sina – cosb

= cosa sina 1 cosb + sina 1 + sinbco–scbosa+ cosa
+ – cosb sina – cosb

= cosa [sina – cosb – sina – cosb] + sinbco–scbosa+ cosa
sin2a – cos2b sina – cosb

= 1 – –2 cosa . cosb + sinbco–scbosa+ cosa
cos2a – 1 + sin2b sina – cosb

= –s2inc2obs–a . cosb + sinbco–scbosa+ cosa
cos2a sina – cosb

= –2 cosa . cosb + cosb . sinb + cosa . cosb + cosa
sin2b – cos2a sina – cosb

= cosb . sinb – cosa . cosb + cosa
sin2b – cos2a sina – cosb

= (sinbc–oscbo(ssain) b(s–incbos+a)cosa) + cosa
sina – cosb

= sinaco–sacosb + cosb
sinb + cosa

 LHS = RHS proved

7. a) (2 – cos2θ) + (1 + 2 cot2θ)= (2 + cot2θ) (2 – sin2θ)

Solution
LHS (2 – cos2θ) + (1 + 2 cot2θ)

= (2 – cos2θ) 1 + 2 . cos2θ
sin2θ 

= 2 – cos2θ (sin2θ + 2 cos2θ)
sin2θ 

=  2 – cos2θ  (sin2θ + 2(1 – sin2θ))
sin2θ sin2θ 

= (2 cosec2θ – cot2θ) (sin2θ + 2 – 2sin2θ)
= [(2(1 + cot2θ) – cot2θ] (2 – sin2θ)
= [2 + 2 cot2θ – cot2θ] (2 – sin2θ)
= (2 + cot2θ) (2 – sin2θ)
 LHS = RHS proved

b) (3 – tan2x) (4 cos2x – 3)= (3 – 4 sin2x) (1 – 3 tan2x)

Solution
LHS (3 – tan2x) (4 cos2x – 3)

= 3 – sin2x  (4 cos2x – 3)
cos2x 

=  3 cos2x – sin2x (4 cos2x – 3)
cos2x 

VedantVaedOapntitoanaOlpMtiaotnhaelmMaattichsemTeaatcichserM'saGnuaidl e~~109 137

= (3(1 – sin2x) – sin2x) 4 cos2x – 3
 cos2x cos2x 

= (3 – 3 sin2x – sin2x) (4 – 3 sec2x)

= (3 – 4 sin2x) (4 – 3(1 + tan2x))

= (3 – 4 sin2x) (4 – 3 – 3 tan2x)
= (3 – 4 sin2x) (1 – 3 tan2x)
 LHS = RHS proved

c) (1 + cotq – cosecq) (1 + tanq + secq) = 2

Solution
LHS (1 + cotq – cosecq) (1 + tanq + secq)
= 1 . (1 + tanq + secq) + cotq(1 + tanq + secq) – cosecq(1 + tanq + secq)
= 1 + tanq + secq + cotq + cotq . tanq + cotq . secq

– cosecq – cosecq . tanq – cosecq . secq

= 1 + (tanq + cotq) + secq + 1 + cosq . 1 – cosecq – 1 . sinq – cosecq . secq
sinq cosq sinq cosq

= 2 +  sinq + cosq  + secq + cosecq – cosecq – secq – cosecq . secq
cosq sinq

= 2 + sin2q + cos2q – cosecq . secq
cosq . sinq

= 2 + 1 sinq – cosecq . secq
cosq .

= 2 + cosecq . sinq – cosecq . secq
= 2
 LHS = RHS proved

8. a) 1 + (cosecA . tanB)2 = 1 + (cotA . sinB)2
1 + (cosecC . tanB)2 (1 + cotC . sinB)2
Solution

LHS 1 + (cosecA . tanB)2
1 + (cosecC . tanB)2

= 11 ++ cosec2A . tanB)2
cosec2C . tan2B

1+ (1 + cot2A) . sin2B
= cos2B

1 +(1 + cot2C) . sin2B
cos2B

cos2B + sin2B + cot2A . sin2B

= cos2B

cos2B + sin2B + cot2C . sin2B

cos2B

= 11 ++ cot2A . sin2B
cot2C . sin2B

138 Vedanta Optional Mathematics Teacher's Guide ~ 9

= 1(1++(ccoottAC . sinB)2
. sinB)2

 LHS = RHS proved

b) 1 + (secA . cotC)2 = 1 + (tanA . cosC)2
1 + (secB . cotC)2 1 + (tanB . cosC)2

Solution

LHS 1 + (secA . cotC)2
1 + (secB . cotC)2

= 1 + sec2A . cot2C
1 + sec2B . cot2C

1+ (1 + tan2A) . cos2C
= sin2C

1 +(1 + tan2B) . cos2C
sin2C

sin2C + cos2C + tan2A . cos2C

= sin2C

sin2C + cos2C + tan2B . cos2C

sin2C

= 1 + tan2A . cos2C
1 + tan2B . cos2C

= 1 + (tanA . cosC)2
1 + (tanB . cosC)2

 LHS = RHS proved

9. a) If 5 cosq + 12 sinq = 13, find the value of tanq

Solution
Given, 5 cosq + 12 sinq = 13
Diving both sides by cosq, we get

5ccoossqq + 12 sinq = 13
cosq cosq

or, 5 + 12 tanq = 13 secq
Squaring both sides

(5 + 12 tanq)2 = (13 secq)2
or, 25 + 120 tanq + 144 tan2q = 169 sec2q
or, 25 + 120 tanq + 144 tan2q = 169 + 169 tan2q
or, 25 tan2q – 120 tanq + 144 = 0
or, (5 tanq)2 – 2.5 tanq . 12 + 122 = 0
or, (5 tanq – 12)2 = 0
or, 5 tanq – 12 = 0

 tanq = 12
5

Vedanta Optional Mathematics Teacher's Guide ~ 9 139

b) If 8 sinq = 4 + cosq, find the value of tanq

Solution

Given, 8 sinq = 4 + cosq
Diving both sides by cosq, we get

8 tanq = 4 secq + 1

or, 8 tanq – 1 = 4 secq

Squaring both sides

(8 tanq – 1)2 = (4 secq)2

or, 64 tan2q – 16 tanq + 1 = 16 sec2q

or, 64 tan2q – 16 tanq + 1 = 16 + 16 tan2q

or, 48 tan2q – 16 tanq – 15 = 0

or, 48 tan2q – 36 tanq + 20 tanq – 15 = 0

or, 12 tanq(4 tanq – 3) + 5(4 tanq – 3) = 0

or, (4 tanq – 3) (12 tanq + 5) = 0

Either 4 tanq – 3 = 0 or, 12 tanq + 5 = 0

4 tanq = 3 tanq = –152

 tanq = 3
5

10. a) If 3 sinq + 4 cosq = 5, prove that: tanq = 3
4
Solution

Given, 3 sinq + 4 cosq = 5

Diving both sides by cosq, we get

3csoisnqq + 4 cosq = 5
cosq cosq

or, 3 tanq + 4 = 5 secq

Squaring both sides

(3 tanq + 4)2 = (5 secq)2

or, 9 tan2q + 24 tanq + 16 = 25 sec2q

or, 9 tan2q + 24 tanq + 16 = 25 + 25 tan2q

or, 16 tan2q – 24 tanq + 9 = 0

or, (4 tanq)2 – 2 . 4 tanq . 3 +32 = 0

or, (4 tanq – 3)2 = 0

or, 4 tanq – 3 = 0

 tanq = 3
4

b) If 4 cos2q + 4 sinq = 5, prove that: sinq = 1
2

140 Vedanta Optional Mathematics Teacher's Guide ~ 9

Solution
Given, 4 cos2q + 4 sinq = 5
4 – 4 sin2q + 4 sinq = 5
or, 4 sin2q – 4 sinq + 1 = 0
or, (2 sinq – 1)2 = 0
or, 2 sinq – 1 = 0

 sinq = 1
2

c) If tanq + cosq = x, show that: sinq = x2 – 1
x2 + 1
Solution

Given, tanq + cosq = x

csoinsqq + 1 = x
cosq

or, sinq + 1 = x . cosq

Squaring both sides

sin2q + 2 sinq + 1 = x2 . cos2q

or, sin2­q + 2 sinq + 1 = x2 – x2 sin2q

or, sin2q + x2 sin2q + 2 sinq + 1 – x2 = 0

or, (1 + x2) sin2q + 2 sinq + (1 – x2) = 0

or, (1 + x2) sin2q + (1 + x2) sinq + (1 – x2) sinq + (1 – x2) = 0

or, (1 + x2) sinq . [sinq + 1] + (1 – x2) (sinq + 1) = 0

or, [(1 + x2) sinq + (1 – x2)] [sinq + 1] = 0

Either (1 + x2) sinq + (1 – x2) sinq + 1 = 0

or, (1 + x2) sinq = x2 – 1 or, sinq = –1

 sinq = x2 – 1 Proved
x2 + 1

d) If sinq – cosq = 0, then show that: cosecq = ± 2

Solution
Given, sinq – cosq = 0
sin­q = cosq
or, sin2q = cos2q
or, sin2q = 1 – sin2q
or, 2 sin2q = 1

or, sinq = ± 1
2

 cosecq = ± 2

Vedanta Optional Mathematics Teacher's Guide ~ 9 141

e) If sinq + cosq = 2 cosq, then prove that: cosq – sinq = 2 sinq

Solution
Given, sinq + cosq = 2 cosq

Squaring bothe sides

(sinq + cosq)2 = ( 2 cosq)2

or, sin2q + 2 sinq . cosq + cos2q = 2 cos2q
or, 1 + 2 sinq . cosq = 2 – 2 sin2q
or, 2 sin2q = 1 – 2 sinq . cosq
or, 2 sin2q = cos2q + sin2q – 2 cosq . sinq
or, 2 sin2q = (cosq – sinq)2

 cosq – sinq = 2 sinq

f) If tanq + sinq = x and tanq – sinq = y, then show that: x2 – y2 = 4 xy

Solution
Here, x2 – y2 = 4 xy

LHS (tanq + sinq)2 – (tanq – sinq)2 = 4 (tanq + sinq) (tanq – sinq)

or, tan2q + 2 tanq . sinq + sin2q – tan2q + 2 tanq . sinq – sin2q = 4 tan2q – sin2q

or, 4 tanq . sinq = 4 sin2q – sin2θ
cos2θ

or, 4 tanq . sinq = 4 sinq 1 – cos2q
cos2θ

 4 tanq . sinq = 4 sinq . tanq (true) proved.

 LHS = RHS proved

g) If 1 + sin2q = 3 sinq . cosq, then prove that: tanq = 1 or 1
2
Solution

Given, 1 + sin2q = 3 sinq . cosq

Dividing bothe sides by cos2q, we get

1 + sin2q = 3 sinq . cosq
cos2q cos2q cos2q

or, sec2q + tan2q = 3 tanq
or, 1 + tan2q + tan2q = 3 tanq
or, 2 tan2q – 3 tanq + 1 = 0
or, 2 tan2q – 2 tanq – tanq + 1 = 0
or, 2 tanq(tanq – 1) – 1(tanq – 1) = 0
or, (2 tanq – 1) (tanq – 1) = 0

142 Vedanta Optional Mathematics Teacher's Guide ~ 9

Either 2 tanq – 1 = 0 tanq – 1

 tanq = 21  tanq = 1

11. If cos4q + cos2q = 1, then show that:

a) tan4q + tan2q = 1

Solution

Given, cos4q + cos2q = 1
or, cos4q = 1 – cos2q
or, cos4q = sin2q

or, cos2q = tan2q

or, 1 + 1 = tan2q
tan2q

 tan4q + tan2q = 1

b) cot4q – cot2q = 1

Solution
Given, tan4q + tan2q = 1
or, 1 + 1 = 1

cot4q cot2q
or, cot2q + 1 = 1

cot4q

 cot4q – cot2q = 1

12. If tanq = 2xy , prove that: sinq = 2xy
x2 – y2 x2 + y2
Solution

Given, tanq = 2xy = p
x2 – y2 h
p
sinq = b

= p
p2 + b2

= 2xy

(2xy)2 + (x2 – y2)2

= 2xy
= 4x2y2 + x4 – 2x2y2 + y4

2xy = 2xy
(x2 + y2)2 x2 + y2

13. If cotq = pq, prove that: p cosq – q sinq = p2 – q2
p cosq + q sinq p2 + q2

Vedanta Optional Mathematics Teacher's Guide ~ 9 143

Solution

Given, cotq = p
q

p cosq – q sinq

Now, p cosq – q sinq = p sinq sinq
p cosq + q sinq cosq + q

sinq

= p cotq – q
p cotq + q

= p × p – q
q

p × p + q
q

= p2 – q2
p2 + q2

?? Questions for practice

1. Prove that:

a) tan2q – tan2a = sec2q – sec2a
b) cot2q – cos2q = cot2q . cos2q

c) 1 c–otsaAnA + 1 sinA = sinA + cosA
– cotA

2. Prove that:

a) ttaannxx + secx – 1 = sinx + 1
– secx + 1 cosx

b) (2 – cos2q) (1 + 2 cot2q) = (2 + cot2q) (2 – sin2q)

c) secq sinq – 1 + cosq – 1 = 1
+ tanq cosecq + cotq

d) (3 – 4 cos2q) (cosec2q – 4 cot2q) = (3 – cot2q) (1 – 4 cos2q)

e) ssiinnqq + cosq + 1 – 1 + sinq – cosq = 2(1 + cosecq)
+ cosq – 1 1 – sinq + cosq

3. a) If tanq = 43, then prove that: 2 sinq + 3 cosq = 18
3 sinq – 2 cosq

b) If tanq = 2xy , then prove that: sina = 2xy
x2 – y2 x2 + y2

144 Vedanta Optional Mathematics Teacher's Guide ~ 9

UNIT

ten Trigonometric Ratios of
Some Standard Angles

1. Objectives Estimated Teaching Hours : 5
SN Level Objectives

i Knowledge (K) To tell trigonometric ratios of standard angles.
ii Understanding (U) To find trigonometric values of simple expressions with standard angles.
iii Application (A) To verify trigonometric identities with standard angles.
iv Higher Ability (Hot) To solve trigonometric equation with t - ratios of standard angles.

2. Teaching Materials
Table with trigonometric ratios of standard angles.

3. Teaching Learning Strategies:
Derive trigonometric ratios of some standard angles 0°, 30°, 45°, 60°, 90° with figures.
Discuss how to write trigonometric ratios of standard angles 0°, 30°, 45°, 60°, 90°.

Notes:
Trigonometric ratios of standard angles (Page 209 text book)

Some solved problems

1. Find the value of sin260° + sin230°
1 – 4 sin60° tan60°
Solution

Here, sin260° + sin230°
1 – 4 sin60° tan60°

 3 2 +  1 2
=  2 2
 
 

1–4. 3 . 3
2
3 1
= 4 + 4

1–6

= 4 ×4(–5)

= –15 = – 1
5

Vedanta Optional Mathematics Teacher's Guide ~ 9 145

2. Prove that: tan230° + 2 sin60° + tan45° – tan60° + cos230° = 25
12
Solution

LHS tan230° + 2 sin60° + tan45° – tan60° + cos230°

=  1 2 + 2 . 3 + 1 – 3 +  3 2
 3 2 2 

= 13 + 3+1– 3 + 3
4

= 31 + 1 + 3
4

= 4 + 12 + 9
12
= 2125

 LHS = RHS proved

3. If a = 60° and b = 30°, verify the following

tan(a – b) = tana – tanb
Solution 1 + tana . tanb

Here, a = 60° and b = 30°

LHS tan(a – b)

= tan(60° – 30°)

= tan30°

= 1
3

RHS tana – tanb
1 + tana . tanb

= 1 +tanta6n06° 0–°t.anta3n03°0°

3– 1
3
=
1
1+ 3. 3

= 3 – 1 . 1
3 2

= 2 . 1
3 2

= 1
3

 LHS = RHS proved

146 Vedanta Optional Mathematics Teacher's Guide ~ 9

4. Solve the following equations:
a) tan245° – cos260° = x . sin45° . cos45° . tan60°

Solution
Here, tan245° – cos260° = x . sin45° . cos45° . tan60°

or, 1 –  1 2 = x . 1. 1. 3
2  2 2

or, 1 – 1 = 3 . x
4 2

or, 4 4– 1 = 3 . x
2

or, 43 = 3 . x
2

 x= 3
2

b) tan60° . cosec60° + 3 x cot30° = sec60° . cosec30°

Solution

Here, tan60° . cosec60° + 3 x cot30° = sec60° . cosec30°

or, 3 . 2 + 3 . x . 3 = 2 . 2
3

or, 2 + 3x = 4

or, 3x = 4 – 2

 x = 2
3

Questions for practice

1. Prove that:

a) sin90° = 3 sin30° – 4 sin330°

b) tan60° = 2 tan30°
1 – tan230°

2. Solve the following equations:

a) sin230° + x cos230° = 2x tan230° + 4 cot260°

b) tan260° + x cot230° = 4x sin260° + cos60° + cos245°
c) tan2 πc – cos2 πc = x sin πc . sec πc . cot πc
43 444

d) 2cos120° – x sin120° . cos180° = x tan150°

e) Sec230° – tan230° + x tan60° = sin2120°

f) tan2150° – sec2120° + x sin260° = 3tan230°

Vedanta Optional Mathematics Teacher's Guide ~ 9 147

UNIT

eleven Trigonometric Ratios
of Any Angle

Compound Angle Estimated Teaching Hours : 20

Symbol used t = trigonometric

1. Objectives

SN Level Objectives

To tell t - ratios of complementary and supplementary angles.

i Knowledge (K) To tell t - ratios of (180° + q), (270° – q), (270° + q), (360° – q)
To tell t - ratios of compound angles.

ii Understanding To solve simple problems involving t - ratios of complementary
(U) and supplementary angles and compound angles.

iii Application (A) To use t - ratios of any angle to prove trigonometric identities.

To solve t - equations with t - ratios of any angle.

iv Higher Ability To derive t - ratios of compound angles.

(HA) To solve harder problems of trigonometric identities with

compound angles.

2. Teaching Materials
Table of t - ratios of
Complementary angels i.e. (90° – q)
Supplementary angles i.e. (180° – q)
(90° + q), (180° + q), (270° – q), (270° + q), (360° – q)
3. Teaching Learning Strategies:
Review the concept of complementary and supplementary angles.
Derive the trigonometric ratios of (90° – q), (90° + q), (180° – q), (180° + q), (270° – q), (270° + q)
Discuss how to find trigonometric ratios of any angles, [(90°n ± q), n∈z+] eg. 1020°, 660°, 750° etc.
Solve some questions from exercise 11.1
Give simple concept of compound angles.
Derive the formulae of trigonometric ratios of compound angles.
Solve some questions from exercise 11.2
After derivation of above each of formula teacher will give at least three examples in each
case.

148 Vedanta Optional Mathematics Teacher's Guide ~ 9

Notes:
1. Table for t - ratios of any angle

Angles T-ratios 90° – q 90° + q 180° – q 180° + q 270° – q 270° + q 360° – q
sine cosq cosq sinq –sinq –cosq –cosq –sinq
cos sinq –sinq –cosq –cosq –sinq sinq cosq
tan cotq –cotq –tanq tanq cotq –cotq –cotq
cosec secq secq –secq –secq
sec cosecq cosecq –cosecq –cosecq cosecq –cosecq
cot tanq –cosecq –secq –secq tanq –tanq secq
–tanq –cotq cotq –cotq

2. Trigonometric Ratios of Compound Angles:

i) sin(A + B) = sinA . cosB + cosA . sinB
sin (A – B) = sinA . cosB – cosA . sinB
ii) cos(A + B) = cosA . cosB – sinA . sinB
cos(A – B) = cosA . cosB + sinA . sinB

iii) tan(A + B) = tanA + tanB
1 – tanA . tanB

tan(A – B) = 1 tanA – tanB
+ tanA . tanB

iv) cot(A + B) = cotA . cotB – 1
cotB + cotA

cot(A – B) = cotA . cotB + 1
v) sin(A + B + C) cotB – cotA

= sinA cosB cosC + cosA sinB cosC + cosA cosB sinC – sinA sinB sinC

vi) cos(A + B + C)

= cosA cosB cosC – sinA sinB cosC – sinA cosB sinC – cosA sinB sinC

vii) tan(A + B + C) = 1 – tanB . tanA + tanB + tanC . tanB
tanC – tanC . tanA – tanA

Some solved problems

1. Find the numerical vlaues of the following:

i) sin(–420°)
Solution
Here, sin(–420°)
= –sin420°
= –sin(4 × 90° + 60°)
= –sin60°

= – 3
2

Vedanta Optional Mathematics Teacher's Guide ~ 9 149

ii) cos570°

Solution

Here, cos570°

= cos(6 × 90° + 30°) = –cos0° = √3
2

2. Prove that: a) sin20° cos70° + cos20° sin70° = 1

Solution

LHS sin20° cos70° + cos20° sin70° Alternately

= sin20° cos(90° – 20°) + cos20° sin(90° – 20°) LHS = sin20° cos70° + cos20° sin70°

= sin20° sin20° + cos20° cos20° = sin (20° + 70°)

= sin220° + cos220° = sin90°

= 1 = 1

 LHS = RHS proved  RHS = RHS proved

b) cot9° cot27° cot45° cot63° cot81° = 0

Solution
LHS cot9° cot27° cot45° cot63° cot81°
= cot9° . cot27° . 1 . cot(90° – 27°) . cot(90° – 9°)
= cot9° . cot27° . tan27° . tan9°
= (tan9° . cot9°) (tan27° . cot27°)
= 1.1
= 1
 LHS = RHS proved

3. Find the numerical value of
a) sin245° – 4 sin260° + 2 cos245° + cos2180° + tan2180° (Corrected page 228, Q.N. 3(f))

Solution
Here, sin245° – 4 sin260° + 2 cos245° + cos2180° + tan2180°

=  1 2 – 4 . 3 2 + 2. 1 2 + (–1)2 + 02
 2  2   2

= 21 – 4 . 3 + 2 . 1 + 1 + 0
4 2

= 21 – 3 + 1 + 1

= 21 – 1
= –12
b) cos290° + cos2120° + cos2135° + cos2150° + cos2180°

Solution
Here, cos290° + cos2120° + cos2135° + cos2150° + cos2180°

150 Vedanta Optional Mathematics Teacher's Guide ~ 9