Shubharambha OPT Mathematics 10 Final for CTP 2077

6. Prove that the following pair of straight lines are parallel to each
other .

(a) 3x + 7 – 4y = 0 and 3x – 4y + 9 = 0
(b) 2x – 3y = 5 and 2x – 3y – 7 = 0
(c) cy – dx = 3 and –3dx + 3cy + 8 = 0
7. (a) Show that the line joining the points (5, 7) and (–2, 3) is parallel to the

line 4x – 7y – 10 = 0.
(b) Prove that the line joining the points (5, –8) and (2, 1) is parallel to the

line joining the points (–1, 3) and (2, –6).

8. Prove that the following pair of straight lines are perpendicular to
each other.

(a) 2x + 2y = 11 and 5x – 5y + 3 = 0
(b) 5x = – 12y and 12x – 5y = 13
(c) px + qy + r = 0 and qx – py + k = 0
9. (a) Prove that the line joining the points P(2, –3) and Q(5, –2) is perpendicular

to the line 3x + y + 6 = 0
(b) Prove that the line joining the points (4, 3) and (8, 9) is perpendicular to

the line joining the points (5, –1) and (2, 1).

10. Find the value of 'm' when
(a) The pair of straight lines x + 3y = 7 and 2x + my + 8 = 0 are parallel.
(b) The pair of straight lines 5x + my – 6 = 0 and 5x – 3y – 8 = 0 are parallel.
(c) The line joining the points (3, 4) and (m – 3, 4) is parallel to the line

joining the points (3, 0) and (4, m).

11. Find the value of 'p' so that the following two straight lines
(a) 4x – py + 8 = 0 and 6x + 21y – 5 = 0 are perpendicular
(b) 5x – 3y + 5 = 0 and px – 7y + 2 = 0 are perpendicular
(c) 4x – 7y + 3 = 0 and 3x + py = 5 are orthogonal.

Section 'C'

12. Find the equation of a straight line,
(a) Passing through the point (5, –4) and parallel to the line 4x – 3y + 5 = 0
(b) Passing through the point (–1, 7) and parallel to the line 5x + 3y – 7 = 0
(c) Passing through the point (–3, –2) and parallel to the line joining the

points (4, –1) and (–1, 2).

13. Find the equation of a straight line
(a) Passing through the point (4, 6) and perpendicular to the line x – 2y = 2.
(b) Passing through the point (3, 4) and orthogonal to the line 5x – 7y = 13.

Infinity Optional Mathematics Book - 10 151

(c) passing through the midpoint of the line joining the points (3, 4) and (–1,
–6) and perpendicular to the line 6x – 7y + 8 = 0.

14. (a) Find the equation of the straight line parallel to the line 5x + 4y = 9 and
making x -intercept is –5.

(b) Find the equation of straight line parallel to the line joining the points
(3, 4) and (7, –1) and making y-intercept –3.

(c) Find the equation of straight line perpendicular to the line 5x – 6y = 8
and making x-intercept is 3.

(d) Find the equation of straight line perpendicular to the line joining the
points (3, –4) and (8, 4) and making y-intercept 5.

15. (a) Find the equation of the straight line passing through the point of
intersection of the lines x + y = 5 and x – y = 1 and parallel to the line 2x
+ 3y = 5.

(b) Find the equation of a line parallel to x – 3y = 4 and passing through the
centroid of the ∆ABC whose vertices are A(3, –4), B(–2, 1) and C(5, 0).

(c) Find the equation of straight line which is perpendicular to the line 3x –
4y + 5 = 0 and passes through the point of intersection of the lines 2x +
y = 5 and x – y = 1.

(d) Find the equation of the line passing through the point of intersection of
the lines x + 2y = 4 and x – 2y = 0 and perpendicular to the line 6x – 7y
+ 8 = 0.

16. (a) Find the equation of straight lines passing through the point (2, 3) and
making angle of 45° with the line x – 3y – 2 = 0.

(b) Find the equation of the straight lines passing through the point (3, –2)
and making angle of 60° with the line 3x + y + 1 = 0.

(c) Find the equation of straight lines passing through origin and making
angle of 60° with the line x + y + 3 = 0.

17. Find the equation of the perpendicular bisector (right bisector) of

the line segment joining the following two points.

(a) (4, –5) and (–8, 9) (b) (3, 5) and (9, 3)

(c) (4, –3) and (8, 5) (d) (4, –2) and (6, 2)

18. (a) If A(2, 3) and C(–6, 5) are the end points of diagonal AC of a square

ABCD, find the equation of other diagonal BD.

(b) B(2, 4) and D(8, 10) are the vertices of a rhombus ABCD, find the equation

of diagonal AC.

19. (a) If the line x + y = 1 passes through the point of intersection of the lines
a b

x + y = 3 and 2x = 3y + 1 and is parallel to the line x – y = 6 then find the

value of 'a' and 'b'.
152 Infinity Optional Mathematics Book - 10

(b) If the line y = mx + b passes through the point of intersection of the lines

x – 2y = 0 and 3x – y = 5 and perpendicular to the line 3x + y + 7 = 0, then

find the values of 'm' and 'b'.

20. (a) A(0, 2) and D(2, 1) are the two ends of a median AD of ∆ABC drawn

from A to BC. Find the equation of a straight line passing through the

centroid and perpendicular to AD.

(b) A(–1, 5) and D(4, 4) are the ends of a median AD of ∆ABC drawn from A

to BC. Find the equation of a straight line passing through the centroid

and perpendicular to AD. C

21. (a) In the adjoining figure, the point D divides the

line segment MN in the ratio of 2:1 and CD

perpendicular to MN. Find the equation of CD. 2 1

M(5, 4) D N(–3, 2)

A(3, 4)

(b) In the given figure, AT perpendicular to MN.

Find the equation of AT.

M (–2, 2) T N(3, –3)

4.2. Equation of a Pair of Straight Lines

Can you write the equation of straight line passing through origin?
Find the product of given pair of straight lines.
(i) 3x + 4y = 0 and x + y = 0
(ii) x – y = 0 and 2x – 3y = 0
(iii) 2x + y = 0 and 5x = y
(iv) 5x – 7y = 0 and 3x + 8y = 0
What are the nature of the product of above equations? Discuss in the

classroom.

Infinity Optional Mathematics Book - 10 153

General equation of second degree:

Let us consider two linear equations
a1x + b1y + c1 = 0 ...........(i)
and a2x + b2y + c2 = 0 ............ (ii)
The product of equation (i) and (ii), we get

(a1x + b1y + c1) (a2x + b2y + c2) = 0
or, a1x (a2x + b2y + c2) + b1y (a2x + b2y + c2) + c1(a2x + b2y + c2) = 0
or, a1a2x2 + a1b2 xy + a1c2x + a2b1xy + b1b2y­2+b1c2y +a2c1x + b2c1y + c1c2 = 0
or, a1a2x­2 + (a1b2 + a2b1)xy + b1b2y­2 + (a1c2 + a2c1) x + (b1c2 + b2c1)y + c1c2= 0 ... (iii)
When, a1a2 = a, b1b2 = b, c1c2 = c, a1b2 + a2b1 = 2h, a1c2 + a2c1 = 2g.
b1c2 + b2c1 = 2f then equation (iii) becomes
ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 is the required general equation of second

degree in x and y. It represents a pair of straight lines.

Homogenous equation of second degree

Let, us consider two equations of straight lines passing through origin,

a1x + b1y = 0 ...... (i) and a2x + b2y = 0 ..... (ii)
Combining equation (i) and (ii), we get

(a1x + b1y) (a2x + b2y) = 0
or, a1x(a2x + b2y) + b1y (a2x + b2y) = 0
or, a1a2x2 + a1b2xy + a2b1xy + b1b2y2 = 0
or, a1a2x2 + (a1b2 + a2b1)xy + b1b2y2 = 0 ....... (iii)
Substituting a1a2 = a, b1b2 = b and a1b2 + a2b1 = 2h in equation (iii), we get
ax2 + 2hxy + by2 = 0 is the required homogenous equation of second degree in

x and y.
Therefore, an equation in x and y in which the sum of the power of x and y in

every term is 2 is called a homogenous equation of second degree.

154 Infinity Optional Mathematics Book - 10

Prove that homogenous equation of second degree
ax2 + 2hxy + by2 = 0 always represent a pair of straight lines and

passing through origin.

The given homogenous equation of second degree is

ax2 + 2hxy + by2 = 0

Multiplying the equation on both sides by a we get

a2x2 + 2ahxy + aby2 = 0

or, a2x2 + 2ahxy + h2y2 – h2y2 + aby2 = 0

or, (ax)2 + 2.ax. hy + (hy)2 – (h2 – ab)y2 = 0

or, (ax + hy)2 – ( h2 – ab. y)2 = 0

or, (ax + hy + h2 – ab.y) (ax + hy – h2 – ab.y) = 0

or, Either, OR, ax + hy – h2 – ab y = 0
ax + hy + h2 – ab.y = 0 or, y(h – h2 – ab) = –ax

or, y(h + h2 – ab) = –ax or, y= –a x .......... (ii)
h – h2 – ab
or, y= –a x ..... (i)
h + h2 – ab

Hence, equation (i) and (ii) are the first degree in x and y and independent of

the constant term and hence represent the pair of lines through the origin.
Hence, ax2 + 2hxy + by2 = 0 always represents a pair of lines through the

origin.

Angle between a pair of lines represented by the equation

ax2 + 2hxy + by2­ = 0. Y

The homogenous equation of second x
degree in x and y is y=m1

ax2 + 2hxy + by2 = 0 y = m 2x
or, by2 + 2hxy + ax2 = 0
Dividing both sides by 'b', we get

y2 + 2 hb.xy + a x2 = 0 ............ (i) q
b O
Y'
X' X

Since, the homogenous equation of second

degree in x and y always represents a

pair of lines passing through origin. Let,

the lines are

y = m1x and y = m2x

Infinity Optional Mathematics Book - 10 155

or, y – m1x = 0 ............. (ii)
and y – m2x = 0 ............. (iii)
Combining equation (ii) and (iii), we get

(y – m1x) (y – m2x) = 0

or, y2 – m2xy – m1xy + m1m2x2 = 0

or, y2 – (m1 + m2) xy + m1m2x2 = 0 ..... (iv)

Now, comparing equation (i) and (iv), we get

m1 + m2 = – 2bh, m1m2 = a
b
We know that,

m1 – m2 = (m1 + m2)2 – 4m1.m2


= – 2h 2 – 4 × a
b b

= 4h2 – 4a
b2 b

= 4h2 – 4ab
b2

= 2 h2 – ab
b

Let, θ be the angle between two lines y = m1x and y = m2x

By formula, tanθ = ± m1 – m2
1 + m1m2
2
b h2 – ab
or, tanθ = ±
a
1 + b

or, tanθ = ± 2 h2 – ab
b b+a

b

or, tanθ = ± 2 h2 – ab
a+b

or, θ = tan-1 ± 2 h2 – ab is the required angle between the pair of lines
a+b

represented by ax2 + 2hxy + by2 = 0

Again, the angle between the pair of lines represented by general equation of
156 Infinity Optional Mathematics Book - 10

second degree ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 is also,

θ = tan-1 ±2 h2 – ab
a+b

Condition for perpendicularity

If the pairs of lines represented by ax2 + 2hxy + by2 = 0 are perpendicular to

each other then, θ = 90°

We have, tanθ = ± 2 h2 – ab
a+b

or, tan90° = ± 2 h2 – ab
a+b

or, csoins9900°° = ± 2 h2 – ab
a+b

or, 01 = ± 2 h2 – ab
a+b

a + b = 0 is the required condition for perpendicular.

Condition for coincident
If the pair of lines represented by ax2 + 2hxy + by2 = 0 are coincident to each

other, then, θ = 0°
We have, tanθ = ± 2 h2 – ab

a+b

or, tan0° = ± 2 h2 – ab
a+b

or, csoins00°° = ± 2 h2 – ab
a+b

or, 01 = ± 2 h2 – ab
a+b

or, ± 2 h2 – ab = 0
or, h2 – ab = 0

Squaring on both sides, we get
h2 – ab = 0
or, h2 = ab is the required condition for coincident.

Can you find the angle between pair of lines represented by the equation
ax2 + 2hxy + by2 = 0 by any other methods ? Discuss in the class.

Infinity Optional Mathematics Book - 10 157

WORKED OUT EXAMPLES

1. Find the single equation represented by the equation 3x + 2y = 0 and
4x – y = 2.

Solution: Here,
The given equations of lines are
3x + 2y = 0 .......... (i)
and 4x – y = 2
or, 4x – y – 2 = 0 .......... (ii)
Combining equation (i) and (ii), we get
(3x + 2y) (4x – y – 2) = 0
or, 3x(4x – y – 2) + 2y (4x – y – 2) = 0
or, 12x2 – 3xy – 6x + 8xy – 2y2 – 4y = 0
or, 12x2 + 5xy – 2y2 – 6x – 4y = 0 is the required single equation of pair of

lines.

2. Find the separate equation of the lines represented by the following
equations

a) x2 – 5xy + 4y2 = 0
b) x2 + 3xy + 2y2 – x – 4y – 6 = 0
c) x2 – 2xycosecθ + y2 = 0
Solution: Here,
a) The given equation is x2 – 5xy + 4y2 = 0
or, x2 – (4 + 1) xy + 4y2 = 0
or, x2 – 4xy – xy + 4y2 = 0
or, x (x – 4y) –y (x – 4y) = 0
or, (x – 4y) (x – y) = 0
Either, x – 4y = 0 .......... (i) or, x – 4y = 0 ........... (ii)
Hence, equation (i) and (ii) are the required separate equations.
b) The given equation of line is
x2 + 3xy + 2y2 – x – 4y – 6 = 0
or, x2 + (3y – 1) x + (2y2 – 4y – 6) = 0 .......... (i)
Comparing equation (i) with ax2 + bx + c = 0, we get

158 Infinity Optional Mathematics Book - 10

a = 1, b = 3y – 1 and c = 2y2 – 4y – 6
By quadratic formula,

x = – b ± b2 – 4ac
2a

or, x= –(3y – 1) ± (3y – 1)2 – 4. 1.(2y2 – 4y – 6)
2×1

or, x = – (3y – 1) ± 9y2 – 6y + 1 – 8y2 + 16y + 24
2

or, x = – (3y – 1) ± y2 + 10y + 25
2

or, x= – (3y – 1) ± (y)2 – 2.y.5 + (5)2
2

or, x= –(3y – 1) ± (y + 5)2
2

or, 2x = –(3y – 1) ±(y + 5)
or, 2x + 3y – 1 = ±(y + 5)
Taking (+ve) sign, we get
2x + 3y – 1 = y + 5
or, 2x + 3y – y = 5 + 1
or, 2x + 2y = 6
x + y – 3 = 0 ........... (ii)
Similarly, taking (–ve) sign we get
2x + 3y – 1 = – (y + 5)
or, 2x + 3y – 1 = – y – 5
or, 2x + 3y + y – 1 + 5 = 0
or, 2x + 4y + 4 = 0
or, 2(x + 2y + 2) = 0
or, x + 2y + 2 = 0 ............. (iii)
Hence, equation (ii) and (iii) are the required separate equations.
(c) The given equation of pair of lines is
x2 – 2xycosecθ + y2 = 0
or, x2 – 2xycosecθ + y2 × 1 = 0
or, x2 – 2xycosecθ + y2(cosec2θ – cot2θ) = 0
or, x2 – 2xycosecθ + y2cosec2θ – y2cot2­ θ = 0

Infinity Optional Mathematics Book - 10 159

or, (x)2 – 2.x.ycosecθ + (ycosecθ)2 – (ycotθ)2 = 0
or, (x – ycosecθ)2 – (ycotθ)2 = 0
or, (x – y cosecθ + ycotθ) (x – ycosecθ – ycotθ) = 0
Either, x– y cosecθ + ycotθ = 0
x – (cosecθ – cotθ)y = 0 ............. (i)
OR, x – y cosecθ – ycotθ = 0
or, x – (cosecθ + cotθ)y = 0 ............. (ii)

Hence, equation (i) and (ii) are required separate equations.

3. Find the angle between the pair of lines represented by

x2 + 5xy + 6y2 = 0

Solution: Here,

The given equation of pair of lines is x2 + 5xy + 6y2 = 0 ..........(i)

Comparing equation (i) with homogenous equation ax2 + 2hxy + by2 = 0

We get,

a = 1, b = 6 and 2h = 5 ⇒ h = 5
2

Let, 'θ' be the angle between pair of lines then by formula,

or, tanθ = ± 2 h2 – ab
a+b

±2 5 2
tanθ = 2
– 1.6

or, 1+6

± 2 25 – 6
1 4
or, tanθ =
+6

± 2 25 – 24
4
or, tanθ = 7

or, tanθ ± 2× 1
= 7 2

tanθ = ± 1
7

160 Infinity Optional Mathematics Book - 10

Taking positive sign, we get Taking negative sign we get

tanθ = 1 tanθ = – 1
7 7

or, θ = tan-1 1 or, θ = tan-1 – 1
7 7

Hence, tan-1 1 and tan-1 – 1 are the required angles between pair of lines.
7 7

4. If the pair of lines represented by an equation kx2 – 5xy – 6y2 = 0 are

perpendicular to each other, find the value of 'k'.

Solution: Here,

The given equation of pair of lines is kx2 – 5xy – 6y2 = 0 ......... (i)

Comparing equation (i) with ax2 + 2hxy + by2 = 0, we get

a = k, 2h = –5 ⇒ h = – 5 and b = – 6
2
Since, the pair of lines represented by the given equation (i) are perpendicular

then
a+b=0
or, k + (–6) = 0
or, k – 6 = 0
∴ k = 6
Hence, k = 6

5. Prove that the lines represented by 9x2 –24xy + 16y2 = 0 are coincident.

Solution: Here,

The given equation of a pair of lines is 9x2 – 24xy + 16y2 = 0 ........ (i)

Comparing equation (i) with ax2 + 2hxy + by2 = 0, we get

a = 9, 2h = – 24 ⇒ h = – 12 and b = 16

h = –12

If the pair of lines represented by the given equation (i) are coincident then

h2 – ab = 0

Now, h2 – ab

= (–12)2 – 9 × 16

= 144 – 144

= 0

Since, h2 – ab = 0, hence, it is proved that the pair of lines represented by

9x2 – 24xy + 6y2 = 0 are coincident.

Infinity Optional Mathematics Book - 10 161

6. Find the single equation of a pair of straight lines through the origin
and perpendicular to the lines represented by x2 – 5xy + 4y2 = 0

Solution: Here,

The given equation of pair of lines is
x2 – 5xy + 4y2 = 0
or, x2 – (4 + 1) xy + 4y2 = 0
or, x2 – 4xy – xy + 4y2 = 0
or, x(x – 4y) – y(x – 4y) = 0
or, (x – 4y) (x – y) = 0
Either, x – 4y = 0 ............(i)
OR, x – y = 0 ............. (ii)
The perpendicular equation of (i) and (ii) are respectively,
4x + y + k1 = 0 ........... (iii)
and x + y + k2 = 0 ............ (iv)
The lines represented by equation (iii) and (iv) passes through origin (0,0)

then. From equation (iii), 4.0 + 0 + k1 = 0
or, k1 = 0
From equation (iv), 0 + 0 + k2 = 0
or, k2 = 0
Substituting k1 = 0 and k2 = 0 in equation (iii) and (iv) respectively. Then,
4x + y = 0 .............. (v)
and x + y = 0 .............. (vi)
Combining equation (v) and (vi), we get
(4x + y) (x + y) = 0
or, 4x(x + y) + y(x + y) = 0
or, 4x2 + 4xy + xy + y2 = 0
or, 4x2 + 5xy + y2 = 0 is the required single equation.

7. Find the single equation of the lines through origin and perpendicular
to the lines represented by ax2 + 2hxy + by2 = 0

Solution: Here,

The given equation of pair of lines is

ax2 + 2hxy + by2 = 0 .......... (i)

Let, y = m1x ........... (ii) and y = m2x .......... (iii) be the separate equations of the
given pair of lines.

Combining equation (ii) and (iii), we get

(y – m1x) (y – m2x) = 0

162 Infinity Optional Mathematics Book - 10

or, y(y – m2x) – m1x(y – m2x) = 0
or, y2 – m2xy – m1xy + m1m2x2 = 0
or, y2 – (m1 + m2) xy + m1m2x2 = 0 ............ (iv)

Dividing given equation (i) on both sides by b, we get

or, abx2 + 2hxy + by2 = 0
b b

or, y2 + 2h xy + a x2 = 0 ......... (v)
b b

Equation (iv) and (v) are identical. So by comparing, we get

– (m1 + m2) = 2h , m1+ m2 = – 2h and m1m2 = – a
b b b

The perpendicular equation of (ii) and (iii) are respectively

m1y + x + k1 = 0 ............ (vi)
and m2y + x + k2 = 0 ............ (vii)
Since, equation (vi) and (vii) passes through (0, 0) then from equation (v) and

(vi)

k1 = 0 and k2 = 0
Substituting k1 = 0 and k2 = 0 in equation (vi) and (vii) we get
m1y + x = 0 .......... (viii)
and m2y + x = 0 .............. (ix)
Combining equation (viii) and (ix), we get

(m1y + x) (m2y + x) = 0

or, m1y (m2y + x) + x(m2y + x) = 0

or, m1m2y2 + m1xy + m2xy + x2 = 0
or, ba y2 + (m1 + m2) xy + x2 = 0
or, ab y2 +
– 2h xy + x2 = 0
b

or, ay2 – 2hxy + bx2 = 0
b

or, bx2 – 2hxy + ay2 = 0 is the required single equation of pair of lines
perpendicular to ax2 + 2hxy + by2 = 0 and passes through origin.

Infinity Optional Mathematics Book - 10 163

Exercise 4.2

Section 'A'

1. (a) What is the angle between pair of lines represented by ax2 + 2hxy +
by2 = 0 ? Write it.

(b) Write any two examples of homogenous quadratic equation.
2. (a) Write the condition for perpendicular (orthogonal) for the pair of lines

represented by homogenous equation ax2 + 2hxy + by2 = 0

(b) Write the condition for coincident for the pair of lines represented by
homogenous equation ax2 + 2hxy + by2 = 0.

(c) What are the pair of lines represented by the equation

(2x + 3y + 1) (4x – y) = 0?
Section 'B'

3. Find the single equation representing the following pair of lines.

(a) 3x – 2y = 0 and x + y = 0

(b) 5x + 6y = 0 and 2x – 7y = 0

(c) 2x + 3y + 2 = 0 and 3x – y = 7

(d) ax = by and bx + ay = 0

(e) (cosecθ – cotθ) x + y = 0 and (cosecθ + cotθ)x + y = 0

(f) y + x(secθ + tanθ) = 0 and y + x(secθ – tanθ) = 0

4. Find the separate equation of the straight lines represented by the
following equation.

(a) 4x2 + 5xy + y2 = 0 (b) 3x2 + 5xy + 2y2 = 0

(c) 3x2 – xy – 4y2 = 0 (d) 33x2 – 44xy + 11y2 = 0

5. Find the separate equation of the lines represented by the given
equations.

(a) x2 + x – y – y2 = 0 (b) x2 – y2 – 3x + 3y = 0

(c) 3x2 – 5xy – 2y2 – x + 2y = 0 (d) x2 – 6xy + 9y2 – x + 3y = 0

(e) xy – 5x – 4y + 20 = 0 (f) abx2 + a2x + b2x + ab = 0

6. Find the angle between the pair of lines represented by the following
equations.

(a) x2 – 3xy – 4y2 = 0 (b) 3x2 + 7xy + 2y2 = 0

(c) x2 – 2xy cosecα + y2 = 0 (d) x2 + 2xysecφ + y2 = 0

(e) x2 + 3xy + 2y2 + 5y + 4x + 3 = 0

164 Infinity Optional Mathematics Book - 10

7. Prove that the lines represented by the following equations are
perpendicular to each other.

(a) 3x2 + 8xy – 3y2 = 0 (b) –5x2 + 24xy + 5y2 = 0

(c) 11x2 + 2xy – 11y2 = 0 (d) (c – d) x2 + 2cdxy + (d – c)y2 = 0

8. Prove that the lines represented by the following equations are
coincident.

(a) x2 – 8xy + 16y2 = 0 (b) 4x2 – 12xy + 9y2 = 0
(c) x2 + 6xy + 9y2 = 0 y2
(d) x2 – xy + 4 = 0

9. Find the value of 'm' when the lines represented by the following

equations are perpendiculr to each other.

(a) mx2 – 5xy – 6y2 = 0 (b) my2 + 8x2 + 12xy = 0

(c) (m – 8)x2 – 9xy + 3my2 = 0 (d) (m + 2)x2 = 4xy – (2m + 5)y2

10. Find the value of 'p' when the lines represented by the following
equations are coincident

( a) 5x2 + 6xy – py2 = 0

(b) (10p – 1)x2 + (5p + 3) xy + (p –1 )y2 = 0

(c) (p + 1)x2 – 4pxy + 2y2 = 0 (d) 9x2 – 24xy+ py2 = 0

Section 'C'
11. (a) Find the equations of two lines represented by the equation

2x2 + 7xy + 3y2 = 0. Also, find the angle between them.

(b) Obtain the separate equations of the pair of lines represented by the

equation x2 + 9xy + 14y2 = 0. Also, find the angle between them.

12. Find the equation of a pair of straight lines represented by the given
equation.

(a) 2x2 – 7xy + 3y2 + 11x – 13y + 12 = 0

(b) 2x2 – 5xy – 3y2 + 3x + 19y – 20 = 0

13. Find the equation of the pair of lines represented by the equation

(a) x2 + 2xysecα + y2 = 0 (b) x2 – 2cotαxy – y2 = 0

(c) x2 + 2xy tanθ – y2 = 0 (d) x2 – 2xy cosecθ + y2 = 0

Also, find the angle between the pair of lines in each above cases.

14. (a) Find the two separate equations when the lines represented by

kx2 + 8xy – 3y2 = 0 are perpendicular to each other.

(b) Find the separate equations when the lines represented by

2x2 – 3xy + ky2 – x + 2y = 0 are perpendicular to each other.

Infinity Optional Mathematics Book - 10 165

(c) Find the two separate when the lines represented by

mx2 + 12xy + 9y2 = 0 are coincident equation.

15. Find the single equation of the pair of straight lines passing through
the origin and perpendicular to the lines represented by the equation.

(a) x2 + 2xy – 3y2 = 0 (b) 6x­2 + 13xy + 6y2 = 0

(c) 2x2 – 7xy + 5y2 = 0 (d) 2x2 – 3xy – 5y2 = 0

16. (a) Find the equation of the straight lines passing through (–1, 4) and

parallel to the pair of lines 5x2 – 8xy – 4y2 = 0

(b) Find the single equation of the lines through (2, 3) and perpendicular to

the lines 3x2 – 8xy + 5y2 = 0

17. (a) If the angle between the lines represented by 2x2 + kxy + 3y2 = 0 is 45°,

find the value of k and then separate equation of lines.

(b) The angle between a pair of straight lines represented by the equation

x2 + xy – ky2 = 0 is 45°. Find the value of k.

18. (a) If α be the acute angle made by the straight lines represented by the

equation x2 + 2xysecθ + y2 = 0, prove that α = θ.

(b) Prove that angle between the lines represented by the equation (x2 + y2)

sin2α = (xcosθ – ysinθ)2 is 2α.



166 Infinity Optional Mathematics Book - 10

4.3 Conic Section

Let us discuss on the following questions. A vertex
(i) What is the name of adjoining figure?
(ii) What is called the base of the given figure? θ Semi-vertical angle
generator

axis

(iii) What is AO and OB called? Radius of base
OB

The angle between vertical height or axis (AO) and Base area

slant height or generator (AB) of cone is called semi-vertical angle. In the figure,

∠BAO = θ is called semi vertical angle. A is called vertex and OB is called radius

of the base.

A conic section (or simply conic) is a curve obtained as the
intersection of the surface of a cone with a plane. There are four
types of conic section they are

(i) circle (ii) ellipse (iii) parabola

(iv) hyperbola

Two equal right circular cone are connecting with each vertex so nappes
formed solid figure is called double mapped right circular cone or
nappes.

If the right circular cone is cut by a plane and perpendicular to the
axis of the cone, or parallel to the base of cone then so formed geometrical figure is
called circle.

α



If a plane intersects a cone at a given angle with the axis B CA
is greater than the semi-vertical angle and less than 90° αq
then the section is called an ellipse. If the given angle
with the axis be 'θ' and semi-vertical angle be α than B' A'
α < θ < 90°.

Infinity Optional Mathematics Book - 10 167

If an intersecting plane, not passing through the B C A

vertex, is parallel to one of the generator of the cone, α P
M S
then the section is called a parabola.
q A'
If the angle between intersecting plane and axis be N

'θ' is equal to semi vertical angle (α) B' R
Q
i.e. ∠AMC = α, ∠PNC = θ
∴ ∠AMC = ∠PNC
i.e. α = θ

If a plane intersects the double right, cone such that

the angle between the axis and the plane be less than the semi-vertical angle, then

the section is called a hyperbola.

B CA

B' C' A'


Exercise 4.3

1. Which types of conic section is formed for the following condition.
(a) When the intersecting plane is parallel to the axis of cone.
(b) When the intersecting plane is perpendicular to the axis of cone.
(c) When the intersecting plane makes any angles with the axis of cone.
(d) When the intersecting plane is parallel to the generating line of cone.
(e) When the intersecting plane is parallel to the base of cone.

2. Which type of conic section are there in the given diagram.
(a) (b)

168 Infinity Optional Mathematics Book - 10

(c) (d)

3. Define the following basic terms used in conic section.

(a) vertex (b) Generator

(c) Semi-vertical angle (d) nappes (e) axis

4. Which type of conic section is formed for the following condition.
(a) If the angle between an intersecting plane and axis is 90°.
(b) If the angle between an intersecting plane and axis is greater than semi-
vertical angle.
(c) If the angle between an intersecting plane and one of the generator of
the cone is 0°.
(d) If the angle between the axis and the intersecting plane is less than the
semi-vertical angle.

5. Label the given diagram.

B CA

P


Infinity Optional Mathematics Book - 10 169

4.4 Circle

If a plane intersects a cone perpendicular to the axis, then the section is called a

circle. B CA
OR

If a right circular cone is cut by an intersecting plane and

parallel to the base of cone then the section is called a

circle. In the adjoining figure, the shaded region is a circle.

We can also define a circle by a locus.

The locus of a moving point which moves so that its B' C' A'
distance from a fixed point is constant is called a circle.

The fixed point is known as the centre and the fixed distance is known as the radius

of the circle. In the given figure, O is called the centre of circle P1 P
(fixed point) and P is a moving point such that OP = radius (r)

units. Let, P1, P2, P3, ..... be the positions of moving point P such P2 O
that OP = OP1 = OP2 = OP3, = ..... = r units. P3
Equation of a circle with centre at origin and radius r units Y
r P(x,y)
Let, origin O(0, 0) be the centre of circle and r be its radius. X
O
Let P(x, y) be any point on the circumference of circle, then

OP = r units.

By using distance formula, X'
OP2 = (x – 0)2 + (y – 0)2

or, r2 = x2 + y2

or, x2 + y2 = r2 Y'
∴ x2 + y2 = r2 ......... (i)

This relation is true for any point on the circumference of circle. Hence, equation (i)

is the required equation of circle when centre is at (0, 0) and radius 'r'.

Equation of circle with centre at (h, k) and radius 'r' units

Let, P(h, k) be the centre and 'r' be the radius of a Y

circle. Let, A(x, y) be any point on the circumference A(x,y)
of the circle, then AP = r. r

By using distance formula X' O P (h,k)
AP2 = (x – h)2 + (y – k)2 X
or, r2 = (x – h)2 + (y – k)2
or, (x – h)2­ + (y – k)2 = r2 .......... (ii)

This relation is true for every point on the Y'
circumference of circle. Hence, equation (ii) is the

required equation of circle when centre is at (h, k) and radius 'r'.

170 Infinity Optional Mathematics Book - 10

General equation of circle

Since, the equation of circle, with centre (h, k) and radius 'r' is
(x – h)2 + (y – k)2 = r2
or, x2 – 2xh + h­2 + y2 – 2yk + k2 – r2 = 0
or, x2­ + y2 – 2hx – 2ky + h2 + k2 – r2 = 0
or, x2 + y2 + 2gx + 2fy + c = 0 ............ (i)
where, g = –h, f = –k and c = h2 + k2 – r2

The equation (i) is the required general equation of circle.

This equation of circle has the following characteristics.

(i) It is second degree in x and y.
(ii) Coefficient of x2 = Coefficient of y2­

(iii) Coefficient of xy = 0
Thus, the general equation of second degree ax2 + 2hxy + by2 + 2gx + 2fy + c

= 0 will represent a circle if a = b and h = 0.

∴ Centre of circle (h, k) = (–g, –f)
Radius of circle (r) = g2 + f2 – c
OR
Radius of circle (r) = h2 + k2 – c

Equation of circle in a diameter form

Let,O be the centre and A(x1, y1) and B(x2, y2) be the two end points of diameter AB
of a circle AQBPA.

Let, P(x, y) be any point on the circumference of the circle, then

Slope of AP (m1) = y – y1 and P(x,y)
x – x1 O B(x2,y2)

Slope of BP (m2) = y – y2 A(x1,y1)
x – x2

Since, APB is the angle in a semi-circle Q
∴ ∠APB = 90° i.e. AP⊥BP so
m1 × m2 = –1

or, xy – y1 × y – y2 =–1
– x1 x – x2

or, (y – y1) (y – y2) = – (x – x1) (x – x2)
or, (x – x1) (x – x2) + (y – y1) (y – y2) = 0 is the required equation of circle in
diameter form.

Infinity Optional Mathematics Book - 10 171

Some necessary condition Y P(h,k)
(i) When a circle touches rk
on x-axis then r = k X' O
Y'
Y X
X
(ii) When a circle touches r (h,k)
on y-axis then r = h
k
X' O h

Y'
Y

(iii) When a circle touches r (h,k)
on both the axes then r = h = k r

k X

X' h

Y'
Y

(iv) When a circle passing r (h,k)
r
through origin then
r2 = h2 + k2 O

X

(v) Let A and B be the centres of two different circles. P rB
When they touch externally at P as shown in figure.
Then, AB = R + r AR



(vi) A and B be the centres of inner and outer circles respectively.

When the two circles touch internally P

Let, PA = r and PB = R then A B

AB = PB – PA = R – r

∴ AB = R – r

(vii) A and B the centres of two different circles. A N RB
When they intersect with each other as in figure. r M

172 Infinity Optional Mathematics Book - 10

Let, AM = r and BN = R then
AB < R + r

(viii) A and B be the centres of two different circles. A P B
When they are disjoint as shown in the figure. Q
Let, AP = r, BQ = R then
AB > R + r

WORKED OUT EXAMPLES

1. Find the equation of circle with centre at (0, 0) and radius is 4 units.

Solution: Here,

Radius (r) = 4 units, centre = (0, 0) 4
The equation of circle is (0,0)
x2 + y2 = r2

or, x2 + y2 = 42

or, x2 + y2 = 16 is the required equation of circle.

2. Find the equation of the circle centre at (4, 1) and radius 5 units.

Solution: Here, Y

Centre (h, k) = (4, 1)

Radius (r) = 5 units 5

The equation of circle is (4,1)

(x – h)2 + (y – k)2 = r2

or, (x – 4)2 + (y – 1)2 = 52 X' O X
or, x2 – 8x + 16 + y2 – 2y + 1 – 25 = 0

or, x2 + y2 – 8x – 2y – 8 = 0 is the required Y'

equation of circle.

3. Find the equation of circle when the coordinates of the two ends of

its diameter are (3, 4) and (2, –7).

Solution: Here,

Let, (3, 4) = (x1, y1), (2, –7) = (x2, y2) (3,4) (2,–7)

Now, the equation of the circle in diameter form is

given by

(x – x1) (x – x2) + (y – y1) (y – y2) = 0

or, (x – 3) (x – 2) + (y – 4) (y + 7) = 0

Infinity Optional Mathematics Book - 10 173

or, x(x – 2) – 3(x – 2) + y(y + 7) – 4(y + 7) = 0
or, x2 – 2x – 3x + 6 + y2 + 7y – 4y – 28 = 0
or, x2 + y2 – 5x + 3y – 22 = 0 is the required equation of circle.

4. Find the equation of circle having centre at (–3, 5) and touching the

y-axis. Y
Solution: Here,

Centre of circle (h, k) = (–3, 5) and the circle touches

the y-axis so radius (r) = |h| = |–3| = 3 units. (–3, 5)
Now, The equation of circle is

(x – h)­2 + (y – k)2 = r2 X' O X

or, (x + 3)2 + (y – 5)2 = 32

or, x2 + 6x + 9 + y2 – 10y + 25 – 9 = 0 Y'
or, x2 + y2 + 6x – 10y + 25 = 0 is the required

equation of circle.

5. Find the centre and the radius of circle whose equations are

(a) x2 + y2 + 4x – 6y + 4 = 0 (b) 9x2 + 9y2 – 36x + 6y = 107

(c) (3x – 5)2 + (3y + 2)2 = 36

Solution: Here,

(a) The given equation of circle is

x2 + y2 + 4x – 6y + 4 = 0 ......... (i)

Comparing equation (i) with general equation of circle

x2 + y2 + 2gx + 2fy + c = 0, we get

2g = 4 2f = – 6 and C = 4

g = 2 f = –3

By formula, centre of circle (h, k) = (–g, –f) = (–2, 3)

and radius of circle (r) = g2 + f2 – c

= (–2)2 + (–3)2 – 4

= 4 + 9 – 4

= 9 = 3 units

(b) The given equation of circle is

9x2 + 9y2 – 36x + 6y = 107

Dividing the equation on both sides by 9, we get

x2 + y2 – 4x + 6y = 107
9 9

or, x2 + y2 – 4x + 2 y – 107 = 0 ........ (i)
3 9

174 Infinity Optional Mathematics Book - 10

Comparing equation (i) with general equation of circle

x2 + y2 + 2gx + 2fy + c = 0, we get

2g = – 4 2f = 32 c = – 107
g = –2 f = 31, 9

∴ By formula,

Centre of circle (h, k) = (–g, –f) = 2, – 1
3

and radius (r) = g2 + f2 – c

= (2)2 + – 1 2 + 107
3 9

= 4 + 1 + 107
9 9

= 36 + 1+ 107 = 144 = 4 units
9 9

(c) The given equation of circle is (3x – 5)2 + (3y + 2)2 = 36

9x2 – 30x + 25 + 9y2 + 12y + 4 – 36 = 0

9x2 + 9y2 – 30x + 12y – 7 = 0

Dividing both sides by 9, we get

or, x2 + y2 – 390 x – 12 y – 7 = 0
9 9

or, x2 + y2 – 130x – 43y – 7 = 0 ........... (i)
9

Comparing equation (i) with general equation of circle

x2 + y2 + 2gx + 2fy + c = 0. We get,

2g = – 10 2f = – 43 and c = – 7
3 f = – 32 9

g = – 53
By formula,

Centre of circle (h, k) = (–g, –f) = 35, 2 and radius (r) = g2 + f2 – c
3

= – 5 2 – 2 2 7
3 3 9
+ +

Infinity Optional Mathematics Book - 10 175

= 25 + 4 + 7
9 9 9

= 25 + 4 + 7
9

= 36 = 2 units
9

6. Find the equation of the circle which touches the x-axis at the point

(4, 0) and passes through the point (3, 1).

Solution: Here,

Let, P(h, k) be the centre of circle and 'r' be its Y
radius. Since, the circle touches the x-axis, so

radius (r) = k and also h = 4 P(h,k)
∴ Centre of circle (h, k) = (4, k)

Now, the equation of circle is X' (3,1) (4,0) X
(x – h)2 + (y – k)2 = r2 O
or, (x – 4)2 + (y – k)2 = k2 ....... (i)
Y'
Since, the circle passes through the point (3, 1)

So, it must satisfy equation (i)

(3 – 4)2 + (1 – k)2 = k2
or, 1 + 1 – 2k + k2 = k2

or, 2 – 2k = 0

∴ k = 1

Substituting k = 1 in equation (i), we get
(x – 4)2­ + (y – 1)2 = 12
or, x2 – 8x + 16 + y2 – 2y + 1 = 1

or, x2 + y2 – 8x – 2y + 16 = 0 is the required equation of circle.

7. Find the equation of circle passing through the point (4, 3) and
concentric with the circle having equation x2 + y2 + 6x – 8y – 11 = 0.

Solution: Here,
The given equation of circle is
x2 + y2 + 6x – 8y – 11 = 0
The equation of circle concentric with the given circle is,
x2 + y2 + 6x – 8y + k = 0 .......... (i)
Since, it passes through the point (4, 3) it should satisfy equation (i) we get
(4)2 + (3)2 + 6 × 4 – 8 × 3 + k = 0
or, 16 + 9 + 24 – 24 + k = 0
∴ k = –25
Substituting k = – 25 in equation (i), we get

176 Infinity Optional Mathematics Book - 10

x2 + y2 + 6x – 8y – 25 = 0
is the required equation of circle which is concentric with the given circle.

8. Find the equation of a circle which passes through the point (2, 3)
and (–1, 2) and has its centre lies on the line 2x – 3y + 1 = 0.

Solution: Here,

Let, P(h, k) be the centre of circle. Since, the centre (h, k) lies on the line

2x – 3y + 1 = 0 so it must satisfy the equation

2x – 3y + 1 = 0 (2,3) (–1,2)
or, 2h – 3k + 1 = 0

∴ h = 3k – 1 ........... (i) P(h,k) 2x – 3y + 1 = 0
2
We have,

The equation of circle is
(x – h)2 + (y – k)2 = r2 .......... (ii)

Since, the circle passes through the points (2, 3) and (–1, 2). So from equation

(ii), we get
(2 – h)2 + (3 – k)2 = r2 .......... (iii)
and (– 1 – h)2 + (2 – k)2 = r2 .............. (iv)

Equating equation (iii) and (iv), we get
(2 – h)2 + (3 – k)2 = (–1 – h)2 + (2 – k)2
or, 4 – 4h + h2 + 9 – 6k + k2 = 1 + 2h + h2 + 4 – 4k + k2

or, 13 – 4h – 6k = 5 + 2h – 4k

or, 13 – 5 – 4h – 2h – 6k + 4k = 0

or, 8 – 6h – 2k = 0

or, 2(4 – 3h – k) = 0

or, 4 – 3h – k = 0

or, –3h = k – 4
k–4
or, h = –3 .......... (v)

Equating equation (i) and (v), we get

3k2– 1 = k–4
–3
or, –9k + 3 = 2k – 8

or, –9k – 2k = –8 –3

or, –11k = –11

∴ k = 1 3k – 1 3 × 1 – 1 2
2 2 2
From equation (i) h = = = = 1

Infinity Optional Mathematics Book - 10 177

∴ Centre of circle (h, k) = (1, 1)
Substituting h = 1 and k = 1 in equation (iii), we get
(2 – 1)2 + (3 – 1)2 = r2
or, 1 + 4 = r2­
or, r2 = 5
∴ r = 5 units
Again, substituting (h, k) = (1, 1) and r = 5 units in equation (ii), we get

(x – 1)2 + (y – 1)2 = ( 5 )2

or, x2 – 2x + 1 + y2­ – 2y + 1 = 5

or, x2 + y2 – 2x – 2y – 3 = 0

Hence, x2 + y2 – 2x – 2y – 3 = 0 is the required equation of circle.

9. Obtain the equation of circle passing through the points (2, –2), (6, 6)

and (5, 7).

Solution: Here,

Let, (h, k) be the centre of circle and 'r' be its radius.

The equation of circle is (x – h)2 + (y – k)2 = r2 ........ (i) (2, –2)
Since, the circle passes through the points (2, –2), (6, 6)

and (5, 7) so must satisfy equation (i) (h,k) (6,6)
∴ (2 – h)2 + (–2 – k)­2 = r2 ....... (ii)
(6 – h)2 + (6 – k)­2 = r2 ......... (iii) (5,7)

and (5 – h)2 + (7 – k)2 = r2 ......... (iv)

Equating equation (ii) and (iii), we get

(2 – h)2 + (–2 – k)2 = (6 – h)2 + (6 – k)2

or, 4 – 4h + h2 + 4 + 4k + k2 = 36 – 12h + h2 + 36 – 12k + k2

or, 8 – 4h + 4k = 72 – 12h – 12k

or, 4k + 12k = 72 – 8 – 12h + 4h

or, 16k = 64 – 8h

or, 16k = 8(8 – h)

or, 2k = 8 – h

∴ h = 8 – 2k .......... (v)

Similarly, equating equation (iii) and (iv) we get

(6 – h)2 + (6 – k)2 = (5 – h)2 + (7 – k)2

178 Infinity Optional Mathematics Book - 10

or, 36 – 12h + h2 + 36 –12k + k2 = 25 – 10h + h2 + 49 – 14k + k2
or, 72 – 12h – 12k = 74 – 10h – 14k
or, 72 – 74 – 12h + 10h – 12k + 14k = 0
or, –2 – 2h + 2k = 0
or, 2(–1 – h + k) = 0
or, – 1 – h + k = 0
∴ h = k – 1 ............. (vi)
Equating equation (v) and (vi), we get
8 – 2k = k – 1
or, 8 + 1 = k + 2k
or, 9 = 3k
k = 3
From equation (vi) h = 3 – 1 = 2
Substituting, h = 2 and k = 3 in equation (ii), we get
(2 – 2)2 + (–2 – 3)2 = r2
or, 0 + 25 = r2
∴ r = 5 units
Substituting (h, k) = (2, 3) and r = 5 units in equation (i) we get
(x – 2)2 + (y – 3)2 = 52
or, x2 – 4x + 4 + y2 – 6y + 9 = 25
x2 + y2 – 4x – 6y + 13 – 25 = 0
or, x2 + y2 – 4x – 6y – 12 = 0
Hence, x2 + y2 – 4x – 6y – 12 = 0 is the required equation of straight line.

Exercise 4.4

Section 'A'

1. (a) What is the equation of circle having centre (0, 0) and radius is 'a' units
? Write it.

(b) Write the equation of circle having centre (p, q) and radius 'r' units.

(c) What is the equation of circle, if two end points A(x1, y1) and B(x2, y2) of
diameter AB are given?

Infinity Optional Mathematics Book - 10 179

(d) Write the centre and radius of circle whose equation is

x2 + y2 + 2gx + 2fy + c = 0

2. (a) Write the equation of circle having centre (0, 0) and radius 2 units.

(b) What is the radius of circle having equation (x – 6)2 + (y – 8)2 = 72?

3. (a) In the given figure, M is the centre of circle and AB is P
a diameter then what is the value of ∠APB.

(b) In the equation of circle (x – h)2 + (y – k)2 = r2. What A M B

does h and k represents ? Write it.
Section 'B'

4. Find the equation of circle, when centre and radius of circle are given
as

S.N. Centre Radius (units)

a. (0, 0) 6

b. (3, 0) 5

c. (0, 4) 4

d. (2, –1) 3

e. (–4, 1) 4

f. (2, 3) 5

g. (a, a) a2

5. Find the equation of the circle with the following conditions.
(a) Centre at (0, 0) and passing through (3, 4)
(b) Centre at (4, – 1) and passing through (–1, 3)
(c) Centre at (4, 2) and passing through origin.
(d) Centre at (2, 3) and passing through (–2, 0).

6. Find the equation of the circle with the following condition.
(a) Centre at (4, 2) and touching x-axis.
(b) Centre at (–2, 5) and touching y-axis.
(c) Radius is 5 units, touching both the positive axes.
(d) Centre lies in the fourth quadrant, touching both the axes having
diameter 5 units.

7. Find the equation of circle when the coordinates of the two ends of

the diameter are:

(a) (3, 2) and (4, – 2) (b) (–7, –4) and (–3, 0)

(c) (3, 6) and (7, – 2) (d) (a, 0) and (0, –a)

8. Find the coordinates of the centre and radius of the circle from the

following equation of circle.

(a) x2 + y2 = 36 (b) (x – 2)2 + (y – 4)2 = 9

180 Infinity Optional Mathematics Book - 10

(c) (x + 2)2 + (y – 3)2 = 50 (d) (x – a)2 + (y – b)2 = c2
(e) (x + 5)2 + y2 = 121 (f) x2 + (y – 3)2 = 12

9. Find the coordinates of the centre and radius of the circle from the

following equation of circle.

(a) x2 + y2 + 4x – 5 = 0 (b) x2 + y2 – 20y + 75 = 0
1
(c) x2 + y2 – 4x – 6y – 12 = 0 (d) x2 + y2 + 5x + 3y – 2 = 0
(e) 6x – 8y – x2 – y2 = 0
(f) x2 + y2 – 8x + 3 = 0

Section 'C'

10. Find the coordinates of the centre and radius of the circle from the

following equation of circle.

(a) 3x2 + 3y2 + 6x – 9y – 3 = 0 (b) 4x2 + 4y2 – 8x + 24y + 32 = 0

(c) 2x2 + 2y2 – 2x + 6y = 45 (d) (3x – 5)2 + (3y + 2)2 = 36

11. (a) If (2, 4) is one end point of a diameter of the circle with equation

x2 + y2 – 2x – 2y = 8, find the coordinates of other end.

(b) The coordinates of an end point of the diameter of circle

x2 + y2­ – 4x – 6y + 11 = 0 is (8, 4), find the coordinates of the other end.

12. (a) The equation of two diameters of a circle passing through the points

(3, 1) are x + 2y – 1 = 0 and 2x – y – 7 = 0. Find the equation of circle.

(b) Find the equation of the circle having the centre as the point of

intersection of the lines x – y = 4 and 2x + 3y + 7 = 0 and passing through

the point (2, 4).

(c) Find the equation of the circle whose equation of two of the diameters

are x + y = 6 and x + 2y = 4 and having diameter 20 units.

(d) A circle has radius 5 units and the equations of its two diameters are

2x – y = 5 and x – 3y + 5 = 0. Find the equation of the circle and show

that it passes through the origin.

13. (a) Find the equation of circle of centre (1, 3) and passing through the point

of intersection of two lines 3x + 2y = 4 and 4x – 3y = 11.

(b) Find the equation of circle of centre (3, 2) which passes through the

midpoint of the line segment joining (5, 1) and (–1, 7).

14. (a) Find the equation of a circle which is concentric with the circle

x2 + y2 – 8x + 12y + 15 = 0 and passing through the point (5, 4).

Infinity Optional Mathematics Book - 10 181

(b) Find the equation of circle concentric with the circle x2 + y2 – 4x + 6y + 4

= 0 and whose radius is 5 units.

15. (a) Find the equation of the circle passing through the points (4, 1) and (6,

5) and having its centre on the line 4x + y = 16.

(b) Find the equation of the circle which passes through the points (3, –2)

and (–2, 0) and its centre lies on the line 2x – y – 3 = 0

16. Find the equation of the circle which passes through the following points.

(a) (5, 7), (6, 6) and (2, –2) (b) (–4, –2), (2, 6) and (2, –2)

(c) (2, –1), (2, 3) and (4, – 1) (d) (1, –1), (3, 1) and (3, –3)

17. (a) Find the equation of circle whose centre is (4, 5) and its passes through

the centre of circle x2 + y2 + 4x + 6y – 12 = 0

(b) Find the equation of circle of center (4, 2) whose radius is twice the

radius of another circle of equation x2 + y2 – 6x – 4y – 12 = 0.

(c) In the given figure, the circle A with A x2 + y2 – 4x + 6y – 12 = 0
centre X passes through the centre B

Y of the circle B. If the equation of

circle B is x2 + y2­ – 4x + 6y – 12 = 0 X Y
and the coordinates of X are (–4, 5), (–4,5)

then find the equation of circle A.

18. (a) Find the equation of circle which touches the x-axis at a point (3, 0) and

passing through the point (2, 1).

(b) Find the equation of circle which touches the y-axi at the point (0, 4) and

passes through the point (3, 1).

19. (a) If the line x + y = 1 cuts a circle x2 + y2 = 1 at two points, find the distance

between the two points.

(b) If the lines x + y = 3 cuts the circle x2 + y2 – 2x – 3 = 0 at two points, find

the distance between the two points.

20. (a) Find the equation of a circle whose centre lies on second quadrant, radius

4 units touching x-axis and equation of a diameter is 2x + y + 2 = 0.

(b) If the diameter of the circle 4x2 + 4y2 + px – 20y – 3 = 0 is 8 units, find

the value of p.

(c) The centre of a circle having equation x2 + y2 – 10x + py + 13 = 0 is

(5, –2). Find the value of P and also the radius of circle.

182 Infinity Optional Mathematics Book - 10

UNIT

5 TRIGONOMETRY

Review

The basic trigonometric formulae which the students have already learned in class

9 are equally important for class 10 also.

A

Basic trigonometric formulae:

1. (i) sinq = hp , cosecq = h ph
p
b h
(ii) cosq = h , secq = b θ

(iii) tanq = p , cotq = b Bb C
b p

2. (i) tanq = sinq = cosq (ii) cotq = cosq = cosecq
cosq cosecq sinq secq

3. (i) sin2q + cos2q = 1 (ii) sec2q – tan2q = 1 (iii) cosec2q – cot2q = 1

Values of Trigonometric ratios of some Standard Angles:

Angles 0º 30º 45º 60º 90º 120º 135º 150º 180º
Ratios

sin 0 1 1 31 31 1 0
cos 2 22 2 22
tan
cosec 1 3 1 1 0 – 1 –1 – 3 –1
sec 2 2 2 2 2 2
cot
011 3 ∞ – 3 –1 – 1 0
3
3

∞2 22 1 2 22 ∞
33

12 2 2 ∞ –2 – 2 – 2 –1
33

∞ 3 1 1 0 – 1 –1 – 3 ∞
33

Infinity Optional Mathematics Book - 10 183

CAST table

(90 + q), (180 – q) q, (90 –q) , (360° + q)
S
A
(sin and cosec + ve) (All + ve)

T C
(180 + q), (270 – q) (cos and sec + ve)
(tan and cot + ve) (270 + q), (360 – q)

Trigonmetircal Ratios of (90º ± q), (180º ± q), (270º ± q ), (360º ± q), and (± q)

Angles
(90º–q ) (90º+q) (180º–q) (180º+ q) (–q) (270º–q) (270º+q) (360º–q)

Ratios

sin cos q cos q sin q –sin q – sin q –cos q –cos q – sin q

cos sin q –sin q –cos q –cos q cos q –sin q sin q cos q

tan cot q –cot q –tan q tan q –tan q cot q –cot q –tan q

cot tan q –tan q –cot q cot q –cot q tan q –tan q –cot q
sec cosec q –cosec q –sec q –sce q sec q –cosec q cosec q sec q
cosec sec q sec q cosec q –cosec q –cosec q –sec q –sec q –cosec q

Compound angles formulae
Sin(A + B) = sinA . cosB + cosA . sinB sin(A ­– B) = sinA . cosB – cosA . sinB

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

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

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

184 Infinity Optional Mathematics Book - 10

5.1 Multiple Angles

Let A be an angle. Then, 2A, 3A, 4A ….. etc. are called multiple angles of A. In this
section, we will discuss about the trigonometrical ratios of angles 2A and 3A in
terms of A. The formulae of multiple angles are derived with the help of compound
angle formulae.
Trigonometric Ratios of angles 2A in terms of A.

1. sin2A = sin(A + A)

= sinA . cosA + cosA . sinA [∵ sin (A + B) = sinA . cosB + cosA . sinB]
\
= 2sinA . cosA
\
sin2A = 2sinA . cosA … (i)
2sinA . cos2A
Sin2A = 2sinA . cosA = cosA = 2tanA
\ sec2A
2tanA
sin2A = 1 + tan2A … (ii)

Sin2A = 2sinA . cosA = 2sin2A . cosA = 2cotA
sinA cosec2A

sin2A = 1 2cotA … (iii)
+ cot2A

2. cos2A = cos(A + A)
= cosA.cosA – sinA.sinA [ ∵ cos (A+B) = cosA.cosB – sinA.sinB]

\ cos2A = cos2A – sin2A … (i)

Cos2A = cos2A – sin2A

= cos2A – (1 – cos2A) = cos2A – 1 + cos2A

\ cos2A = 2cos2A – 1 … (ii)

Cos2A = cos2A – sin2A

= 1 – sin2A – sin2A

\ cos2A = 1 – 2sin2A … (iii)

Cos2A = 2cos2A – 1

= 2 – 1
sec2A

= 1 + 2 – 1
tan2A
2– 1 – tan2A
= 1 + tan2A

\ cos2A = 1 – tan2A … (iv)
1 + tan2A

Cos2A = 1 – 2sin2A
2
= 1– cosec2A

Infinity Optional Mathematics Book - 10 185

= 1– 1 2
+cot2A

= 1 + cot2A – 2
1 + cot2A

\ cos2A = cot2A – 1 .... (v)
cot2A + 1

3. Tan2A = tan(A + A)

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

\ tan2A = 1 2tanA
– tan2A

4. cot2A = cot(A + A)

cotA × cotA – 1 cotA × cotB – 1
cotA + cotA cotA + cotB
[ ]= ∵ =
cot (A+B)

\ cot2A = cot2A – 1
2cotA

Trigonometric Ratios of angles 3A interms of A

1. sin3A = sin(2A + A)

= sin2A . cosA + cos2A . sinA
= 2sinA . cosA . cosA + (1 – 2sin2A) . sinA
= 2sinA (1 – sin2A) + sinA – 2sin3A

= 2sinA – 2sin3A + sinA – 2sin3A

\ sin3A = 3sinA – 4sin3A

2. cos3A = cos(2A + A)

= cos2A . cosA – sin2A . sinA

= (2cos2A – 1)cosA – 2sinA . cosA . sinA

= 2cos3A – cosA – 2(1 – cos2A) . cosA

= 2cos3A – cosA ­– 2cosA + 2cos3A

\ cos3A = 4cos3A – 3cosA

3. tan3A = tan(2A + A)

= tan2A + tanA
1 – tan2A . tanA

1 2tanA + tanA
– tan2A
=
2tanA
1 – 1 – tan2A . tanA

186 Infinity Optional Mathematics Book - 10

= 2tanA + tanA – tan3A
\ 1 – tan2A – 2tan2A

3tanA – tan3A
tan2A = 1 – 3tan2A

4. cot3A = cot3A – 3cotA
3cot2A – 1

Multiple angles formulae

1. sin2A = 2sinA.cosA 2. sin2A = 1 2tanA
+ tan2A

3. sin2A = 1 2+ccoottA2A 4. cos2A = cos2A – sin2A

5. cos2A = 2cos2A – 1 6. cos2A = 1 – 2sin2A

7. cos2A = 1 – ttaann22AA 8. cos2A = cot2A – 1
1 + cot2A + 1

9. 2cos2A = 1 + cos2A 10. 2sin2A = 1 – cos2A

11. tan2A = 1 2–ttaannA2A 12. cot2A = cot2A – 1
2cotA

13. sin3A = 3sinA – 4sin3A 14. cos3A = 4cos3A – 3cosA
15. tan3A = 3t1an–A3t–atna2nA3A
16. cot3A = cot3A – 3cotA
3cot2A – 1

WORKED OUT EXAMPLES

1. If sinq = 54, find the value of sin2q, cos2q and tan2q.
Solution :
4
Here, sinq = 5

∴ cosq = 1 – sin2q = 1– 4 2 = 35.
5

Now, sin2q = 2sinq . cosq = 2 × 4 × 3 = 24
5 5 25

Cos2q = 1 ­– 2sin2q

= 1 ­– 2 × 42
5

Infinity Optional Mathematics Book - 10 187

= 1 – 32 = –7
25 25

24

Tan2q = sin2q = 25 = – 24
cos2q –7 7

2. If cos2A = 1251, show that 2co5sA = 6
52

Solution :

Here , cos2A = 11
25

or, 2cos2A ­– 1 = 11
25

or, 2cos2A ­ = 11 + 1
25

or, cos2A = 36
50

or, cosA = 36
50

∴ cosA = 6
52

3: Prove that :

(a) 1 +cossi2nA2A = 1 – tanA = cosA – sinA
1 + tanA cosA + sinA

(b) 11 +– cos2 q + sin2 q = tan q
cos2 q + sin2 q

(c) 1 + tanα. tan2α = sec2α.

Solution:

(a) L.H.S. = 1 cos2A
+ sin2A

1 – tan2A

= 1 + tan2A
2tanA
1 + 1 + tan2A

= 1 + 1 – tan2A
tan2A + 2tanA

188 Infinity Optional Mathematics Book - 10

= 1 – tan2A
(1 + tanA)2

= 1 – ttaannAA
1 +

= Middle part (M.P.) proved.

11 – tanA 1– sinA cosA – sinA
+ tanA cos A cosA + sinA
Again, = =
sinA
1 + cosA

\ L.H.S = M.P. = R.H.S. proved.

(b) L.H.S. = 1 – cos2θ + sin2θ
1 + cos2θ + sin2θ

= 2sin2θ + 2sinθ . cosθ = 2sinθ(sinθ + cosθ) = tanq= R.H.S. proved.
2cos2θ + 2cosθ . sinθ 2cosθ(cosθ + sinθ)

(c) L.H.S. = 1 + tanα. Tan2α

= 1 + sin . csoins22
cos

= cos2 . cos + sin2. sin
cos . cos2

= cos(2α – ) = cos = sec2α. = R.H.S. proved
cos . cos2 cosα . cos2α

4. Express cos5q in terms of cosq

Solution:

cos5q = cos(2q + 3q)

= cos2q . cos3q – sin2q . sin3q

= (2cos2q – 1) (4cos3q – 3 cosq) – 2sinq . cosq (3sinq – 4sin3q)

= 8cos5q – 6cos3q – 4cos3q + 3cosq – 6sin2q . cosq + 8sin4q . cosq

= 8cos5q – 10cos3q + 3cosq – 6(1 – cos2q) . cosq + 8cosq (1 – cos2q)2

= 8cos5q – 10cos3q + 3cosq – 6cosq + 6cos3q + 8cosq (1 –2cos2q + cos4q)

= 8cos5q – 4cos3q – 3cosq + 8cosq – 16cos3q + 8cos5q

= 16cos5q – 20cos3q + 5cosq

5. If cosA = 1 a + 1 show that:
2 a
1 1
(a) cos2A = 2 a2 + 1 (b) Cos3A = 2 a3 + 1
a2 a3
Solution :

(a) cos2A = 2 cos2A – 1

Infinity Optional Mathematics Book - 10 189

=2× 1 a + 1 2
2 a
–1

= 2 × 1 a + 1 2 1
\ 4 a
–

= 1 a2 + 2a . 1 + 1 –1
2 a a2
1
= 1 a2 + 1 + 2a2 – 1
2
1
cos2A = 1 a2 + a2
2

(b) Cos3A = 4cos3A – 3cosA

1 a + 1 3 1 1
2 a 2 a
=4× – 3 × a +

= 4 × 1 a + 1 3 3 a + 1
8 a 2 a
–

= 1 a3 + 3a . 1 1 + 1 – 3 a+ 1
2 a a+ a a3 2 a

= 1 a3 + 3 a + 1 + 1 . 1 – 3 a + 1
2 2 a 2 a3 2 a

\ cos3A = 1 a3 + 1
2 a3

6. If 2tanA = 3tanB, then prove that
sin2B

tan (A – B) = 5 – cos2B
Solution :

Here, 2tanA = 3tanB

or, tanA = 3 tanB.
2

L.H.S. = tan (A – B)

= 1 tanA – tanB
+ tanA . tanB

3 tanB – tanB 3tanB – 2tanB
2 2 + 3tan2B
= 3 =
1 2
+ tanB . tanB

sinB

= tanB = cosB = sinB . cosB = sinB . cosB
2 + 3tan2B sin2B 2cos2B + 2sin2B 2cos2B + 3 – 3cos2B
2 + 3 cos2B
sinB . cosB
= 3 – cos2B

190 Infinity Optional Mathematics Book - 10

= 2sinB . cosB = 6 – sin2B = 6 – sin2B = 5 sin2B
6 – 2cos2B (1 + cos2B) 1 – cos2B – cos2B

7. Prove the following :

(a) cosθ – 1 + sin2θ = tan q
sinq + 1 + sin2θ

(b) (1 + sin2A + cos2A)2 = 4cos2A (1 + sin2A)

(c) tan(B + 45°) + tan(B – 45°) = 2 tan2B

(d) cos6A – sin6A = 1 cos 2A(4 – sin22A)
4

(e) cos4A = 1 (3 + 4cos2A + cos4A)
8

(f) cosec2q + cot4q = cotq – cosec4q

(g) 3 cosec20º – sec20º = 4

(h) cos3α × cos3α + sin3α . sin3α = cos32α

(i) (2cosq + 1) (2cosq – 1) (2cos2q – 1) = 2cos4q + 1
Solution:

(a) L.H.S. = cosθ – 1 + sin2θ
sinθ – 1 + sin2θ

=csoinsθθ – sin2θ + cos2θ + 2sinθ . cosθ
– sin2θ + cos2θ + 2sinθ . cosθ

= cosθ – (sinθ + cosθ)2
sinθ – (sinθ + cosθ)2

cosθ – sinθ - cosθ
= sinθ – sinθ – cosθ

= sinθ = tanq = R.H.S. proved.
cosθ

(b) L.H.S. = (1 + sin2A + cos2A)2

= (1 + cos2A + sin2A)2

= (2 cos2A + 2sinA.cosA)2

= [2cosA (cosA + sinA)2

= 4cos2A (cos2A + 2cosA . sinA + sin2A)

= 4cos2A (1 + sin2A)

= R.H.S. Proved.

Infinity Optional Mathematics Book - 10 191

(c) L.H.S. = tan(B + 45°) + tan(B – 45°)

= tanB + tan45° + tanB – tan45°
1 – tanB . tan45° 1 + tanB . tan45°

= tanB + 1 tanB – 1 (tanB + 1)2 – (1 –tanB)2
1 – tanB + 1 + tanB = (1 + tan B) (1 – tan B)

= tan2B + 2tanB + 1 – tan2B + 2tanB – 1
1 – tan2B

= 2 × 2 tanB = 2tan2B
1 – tan2B

= R.H.S. proved

(d) L.H.S. = cos6A – sin6A

= (cos2A)3 – (sin2A)3

= (cos2A – sin2A) (cos4A + cos2A . sin2A + sin4A)

= cos2A [(cos2A + sin2A)2 – 2cos2A . sin2A + cos2A.sin2A]

= cos2A (1 – cos2A . sin2A)

= cos2A 1 – 14sin22A

= 1 cos2A (4 – sin22A)
4

= R.H.S. proved

(e) R.H.S. = 1 (3 + 4cos2A + cos4A)
8

= 1 [3 + 4(2cos2A – 1) + 2cos22A – 1]
8

= 1 [3 + 8cos2A – 4 + 2(2cos2A – 1)2 – 1]
8

= 1 (8cos2A – 2 + 8cos4A – 8cos2A + 2)
8

= 1 × 8cos4A
8

= cos4A = L.H.S. proved.

(f) L.H.S. = cosec2q + cot4q

= 1 + cos4θ
sin2θ sin4θ

192 Infinity Optional Mathematics Book - 10

= 1 + cos4θ
sin2θ 2sin2θ . cos2θ

= 2cos2θ + 2cos22θ – 1
2sin2θ . cos2θ

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

= 2cos2θ . 2cos2θ – 1 cos2θ
2sin2θ . cos2θ 2sin2θ .

= 2cos2θ – 1
2sinθ . cosθ sin4θ

= cotq – cosec4q

= R.H.S. proved

(g) L.H.S. = 3 cosec20º – sec20º

= 3 – 1
sin 20° cos20°

= 3 cos 20° – sin20°
sin 20° . cos20°

3 cos20° – 1 sin20°
2 2
=
1
2 sin20° . cos20°

sin60° . cos20°– cos60° . sin20° = 4 sin (60°–20°)
= 1 × 1 × (2sin 20° . cos20°) sin 40°
22

= 4sin40° = 4 = R.H.S. proved.
sin40°

(h) L.H.S. = cos3α . cos3α + sin3α . sin3α

= cos3α (4cos3α – 3cosα) + sin3α (3sinα – 4sin3α)

= 4cos6α – 3cos4α + 3sin4α – 4sin6α

= 4(cos6α – sin6α) – 3(cos4α – sin4α)

=4[cos2α–sin2α)3+3cos2α.sin2α(cos2α–sin2α)]­

–3(cos2α–sin2α) (cos2α+sin2α)

= 4(cos32α + 3cos2α . sin2α . cos2α) – 3cos2α

= 4cos32α + 3 . (2sinα . cosα)2. cos2α – 3cos2α

= 4cos32α + 3sin22α . cos2α – 3cos2α

Infinity Optional Mathematics Book - 10 193

= 4cos32α + 3cos2α (1 – cos22α) – 3cos2α

= 4cos32α + 3cos2α – 3cos32 α – 3cos2 α

= cos32α

= R.H.s. proved.

(i) L.H.S. = (2cosq + 1) (2cosq – 1) (2cos2q – 1)

= (4cos2q – 1) (2cos2q – 1)

= [2(2cos2q) – 1] (2cos2q – 1)

= [2(1 + cos2q) – 1] (2cos2q – 1)

= (2 + 2cos2q – 1) (2cos2q – 1)

= (2cos2q + 1) (2cos2q – 1)

= 4cos22q – 1

= 2(2cos22q) – 1

= 2(1 + cos4q) – 1

= 2 +2 cos4q – 1

= 2 cos4q + 1

= R.H.S. proved.

Exercise 5.1

1. (a) Find the value of sin2A, cos2A and tan2A when

(i) sinA = 3 (ii) cosA = 153 (iii) tanA = 3
5 4

(b) Find the value of sin3A, cos3A and tan3A when

(i) sinA = 1 (ii) cosA = 1 (iii) tanA = 1
2 2 3

2. (a) If cos2A = 7 then show that sinA = 3
25 5

(b) If cos2q = – 119 then show that cosq = 5
169 13

(c) If tan2q = 24 then show that tanq = 3
7 4

194 Infinity Optional Mathematics Book - 10

3. Using formulae of sin2A, cos2A and tan2A, prove the followings:

(a) sinA = ± 1 – cos2A (b) cosA = ± 1 + cos2A
2 2

(c) tanA = ± 1 – cos2A (d) cotA = ± 1 + cos2A
1 + cos2A 1 – cos2A

(e) sin2A = 1 2cotA (f) sin2A = 2tanA
+ cot2A 1 + tan2A

(g) cos2A = 1 – tan2A (h) cos2A = cot2A – 1
1 + tan2A cot2A + 1

4. Prove the following identities:

(a) 1 +sinco2sA2A = cotA (b) 1 – cos2A = tanA
sin2A

(c) 11 +– cos2A = tan2A (d) 1 +sinco2sA2A = tanA
cos2A

(e) 1 –sinco2sA2A = cotA

(f) 1 c–ossi2nθ2θ = 1 – tanθ = cosθ – sinθ
1 + tanθ cosθ + sinθ

(g) 1 +cossi2nA2A = 1 – tanA = cosA – sinA
1 + tanA cosA + sinA

(h) 1 +cossi2nA2A = cotA + 1 = cosA + sinA
cotA – 1 cosA – sinA

(i) cosθc o–sθsinθ – cosθ = tan2θ (j) tanq + cotq = 2 cosec 2q
cosθ + sinθ

(k) tanq – cotq = – 2cot2q

(l) tan2A + sin2A = 1 4–ttaannA4A (m) ssiinn3AA – cos3A = 2
cosA

(n) sec2q – tan2q = cosθ – sinθ (o) cos4A – sin4A = cos2A
cosθ + sinθ

(p) ccoottAA – tanA = cos2A (q) sin3A + cos3A = 1 – 1 sin2A
+ tanA sinA + cosA 2

(r) (sinq + cosq)2 = 1 + sin2q

(s) ccoossθθ +– sinθ – cosθ – sinθ = 2tan2q (t) 11 – cos2A + sin2A = tanA
sinθ cosθ + sinθ + cos2A + sin2A

(u) 11 ++ sin2A + cos2A = cotA (v) sinθ + sin2θ = tanq
sin2A – cos2A 1 + cos2θ + cosθ

Infinity Optional Mathematics Book - 10 195

(w) cosec2A – sec2A = 4cosec2A . cot2A (x) 1 + tanA . tan2A= sec2A
(y) (1 + tanq – secq) (1 + tanq + secq) = sin2q . sec2q
(z) sin8q = 8sinq . cosq . cos2q . cos4q

5. Prove the followings :

(a) cos2(45° – q) ­– sin2(45° – q) = sin2q (b) 1 – tan2 c – A
1 + tan2 4
c = sin2A
4
– A

(c) 11 +– tan2(45° – A) = cosec2A (d) 2sin2 (45° – q) = 1 – sin2q
tan2(45° – A)

(e) 2cos2 (45° – q) – 1 = sin2q

6. Prove the following identities:

(a) (sinq + cosq)2 – (sinq – cosq)2 = 2sin2q (b) cosA – 1 + sin2A = tanA
sinA – 1 + sin2A

(c) 11 +– sin2A = 1 + tanA 2 (d) 1 c–ossi2nA2A = tan(45° – A)
sin2A 1 – tanA

(e) (1 + sin2q + cos2q)2 = 4cos2q (1 + sin2q)

(f) (1 – sin2q + cos2q)2 = 4cos2q (1– sin2q)

(g) 1 + cos22q = 2(cos4q + sin4q) (h) (sec2A + 1) sec2 A– 1 = tan2A

(i) tan3θ1+ tanθ – 1 cotθ = cot4q
cot3θ +

(j) tan3θ1– tanθ – 1 cotθ = cot2q
cot3θ –

7. Prove the following identities :

(a) tan(A + 45º) + tan(A – 45º) = 2tan2A

(b) cot(A+ 45º) – tan(A – 45º) = 2cos2A
1 + sin2A

(c) tan(A + 45º) – tan(A – 45º) = 2sec2A

(d) cos6q + sin6q = 41(1 + 3cos22q) = 18(5 + 3cos4q)

(e) cos6q – sin6q = (2cos2q – 1) 1 – 1 sin22θ
4

196 Infinity Optional Mathematics Book - 10

(f) cos8q + sin8q = 1 – sin22q + 1 sin42q (g) 2 + 2 + (2 + 2cos8A) = 2cosA
8

(h) sin4A = 1 (3 – 4cos2A + cos4A) (i) cos4A = 1 (3 + 4cos2A + cos4A)
8 8

(j) cos4q = 1 – 8sin²q + 8sin4q (k) cos3θc o–sθcos3θ + sin3θ + sin3θ = 3
sinθ

8. Prove the following identities :

(a) cosec2q + cot4q = cotq – cosec4q (b) sec8θ – 1 = tan8θ
sec4θ – 1 tan2θ

(c) sin230° – 1 = 4 (d) cosec10º – 3 sec10º = 4
cos20°

(e) 4(cos310º + sin320º) = 3(cos10º + sin20º)

(f) cos320º+sin350º = 3 (cos20º + sin50º)
4

(g) sinA . sin(60° – A) . sin(60° + A) = 14sin3A

(h) 4cosq . cos(60° + q) . sin(30° + q) = cos3q

(i) cosA . cos(120° – A) . cos (120° + A) = 1 cos3A
4

(j) tanA + 2tan2A + 4tan4A + 8cot8A = cotA

(k) 4cos3q . sin3q + 4sin3q . cos3q = 3sin4q

(l) cos3α . cos3 α + sin3 α . sin3α = cos32α

(m) cos2 q + sin2 q . cos2β = cos2β + sin2β . cos2 q

(n) cos2 (q – 120º) + cos2 q + cos2 (q + 120º) = 3
2

(o) sin2 q + sin2(q – 120º) + sin2 (q + 120º) = 3
2

(p) sin2A – cos2A . cos2B = sin2B – cos2B . cos2A

(q) c2ossin3AA + 2sin3A + 2sin9A = tan27A – tanA
cos9A cos27A

(r) cos3q + cos3(120° + q) + cos3 (240° + q) = 3 cos 3q
4

9. Express

(a) sin3q in terms of sinq (b) cos 3q in terms of cos q

(c) cot 3q in terms of cotq (d) tan 3q in terms of tanq

Infinity Optional Mathematics Book - 10 197

(e) cos4q in terms of cosq

10. (a) If cosA = 1 = x + 1 , show that
2 x

(i) cos 2A = 1 x2 + 1 (ii) cos 3A = 1 x3 + 1
2 x2 2 x3

(b) If sinA = 1 a + 1 , show that
2 a

(i) cos 2A = – 1 a2 + 1 (ii) sin 3A = –1 a3 + 1
2 a2 2 a3

11. (a) If 2tanA = 3tanB, then prove that

(i) tan (A – B) = 5 sin2B (ii) tan (A + B) = 5sin2B 1
– cos2B 5cos2B –

(b) If tana = ba, show that a cos2a + bsin2a = a.

(c) If tanq = 1 and tanb = 31, prove that cos2q = sin4b
7
1 – tan4
(d) If tanq = sec2a, then show that sin2q = 1 + tan4

198 Infinity Optional Mathematics Book - 10

5.2 Sub-Multiple Angles

Let A be an angle. Then A2 , A3 , A ... etc. are called sub-multiple angles of A. In this
4 A and
section, we will discuss about the trigonometric ratios of angle A interms of 2
A3 .
Trigonometric Ratios of angles A in terms of A2 .

1. sinA = sin A + A
2 2

= sinA2 . cosA2 + cos A . sinA2 [∵sin(A + B) = sinA . cosB + coA . sinB]
2

= 2sinA2 . cosA2 .

\ sinA = 2sinA2 . cosA2

sinA = 2sinA2 . cosA2 = 2tanA2 . cos2A2 = 2tanA2
cosA2 sec2A2

\ sinA = 2tanA2
1 + tan2A2

2. cosA = cos A + A
2 2

= cosA2 . cosA2 – sinA2 . sin A
2

\ cosA = cos2 A – sin2 A … (i)
2 2

CosA = cos2 A2 – sin2 A
2

= cos2A2 – 1 – cos2 A = cos2 A – 1 + cos2 A
2 2 2

\ cosA = 2cos2 A – 1 … (ii)
2

CosA = cos2 A – sin2 A
2 2

= 1 – sin2 A – sin2 A
2 2

Infinity Optional Mathematics Book - 10 199

\ cosA = 1 – sin2A2 … (iii)

CosA = 2cos2 A –1
= 2 2 –1

sec2 A
2

= 2 –1

1 + tan2 A
2

= A
2 – 1 – tan2 2

1 + tan2 A
2

A
1 – tan2 2
\ cosA = ... (iv)
A
1 + tan2 2

CosA = 1 – 2sin2 A
2

= 1 – 2

cosec2 A
2
= 1 – 2
A
1 + cot2 2

A
1 + cot2 2 – 2
=
A
1 + cot2 2

\ cosA = cosA = A – 1 … (v)
cot2 2
+ 1
cot2 A
2

AA
A A tan 2 + tan 2
3. tanA = tan 2 + 2 = AA

1 – tan 2 . tan 2

A
2tan 2
\ tanA =
A
1 – tan2 2

4. cotA = cot A + A
2 2

200 Infinity Optional Mathematics Book - 10