Page 37 - mathsvol1ch1to3ans

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10. If cot θ (1 + sin θ) = 4m and cot θ (1 − sin θ) = 4n, then prove that m − n 2 2 = mn.
Solution:
4m = cot θ + cos θ

4n = cot θ + cos θ

4(m + n) = 2 cot θ

4(m − n) = 2 cos θ

2(m + n) = cot θ
sec θ = Not For Sale - Veeraragavan C S veeraa1729@gmail.com
2(m − n) = cot θ
2
2
4(m − n ) = cot θ cos θ
2
2
2
2
2
cos θ cos θ cos θ(1 − sin θ) cos θ
2
2
2
2
2
16(m − n ) = cot θ cos θ = = = − cos θ
2
2
2
sin θ sin θ sin θ
2
2
= cot θ − cos θ
= 16mn
3
2
2
2 2
3
11. If cosec θ − sin θ = a and sec θ − cos θ = b , then prove that a b (a + b ) = 1.
Solution:
2
2
1 1 − sin θ cos θ
3
a = cosecθ − sin θ = − sin θ = =
sin θ sin θ sin θ
2 4

2
cos θ 3 cos 3 θ
2
a = = (1)
sin θ 2
sin 3 θ
2
2
1 1 − cos θ sin θ
3
b = secθ − cos θ = − cos θ = =
cos θ cos θ cos θ
2 4

2
sin θ 3 sin 3 θ
2
b = = (2)
cos θ 2
cos 3 θ
4 4

cos 3 θ sin 3 θ
2 2
a b = (1) × (2)
2
2

3 3 θ
2
2 θ cos
sin
= cos 3 θ sin 3 θ (3)
4 4 4 2 4 2

cos 3 θ sin 3 θ cos 3 θ cos 3 θ + sin 3 θ sin 3 θ 1
2
2
a + b = + = =
2
2
2
2
2
2

sin 3 θ cos 3 θ sin 3 θ cos 3 θ sin 3 θ cos 3 θ
2 2
2
2
Hence a b (a + b ) = 1.
12. Eliminate θ from the equations a sec θ − c tan θ = b and b sec θ + d tan θ = c.
Solution:Solving both equations for sec θ and tan θ we have
bd + c 2 ac − b 2
and tan θ = .
ad + bc ad + bc
2 2 2 2
bd + c ac − b
2
2
Since sec θ − tan θ = 1 we have − = 1.
ad + bc ad + bc
2 2 2 2 2
Simplifying, we have, (bd + c ) = (ac − b ) + (ad + bc) .
Exercise 3.2:
1. Express each of the following angles in radian measure: