Page 47 - mathsvol1ch1to3ans

47

Solution:
1
cot(A + B) =
tan(A + B)
1 − tan A tan B
=
tan A + tan B
Multiplying both Numerator and Denominator by cot A cot B
cot A cot B − 1
We get
cot B + cot A
n 1
22. If tan x = and tan y = , find tan(x + y).
Not For Sale - Veeraragavan C S veeraa1729@gmail.com
n + 1 2n + 1
Solution:
tan(x) + tan(y)
tan(x + y) =
1 − tan(x) tan(y)
n 1
+
n + 1 2n + 1
= n
1 −
(n + 1)(2n + 1)
n(2n + 1) + n + 1
=
(n + 1)(2n + 1) − n
2
2n + 2n + 1
=
2
2n + 2n + 1
= 1

π 3π

23. Prove that tan + θ tan + θ = −1.
4 4
Solution:

π 3π (1 + tan θ) (−1 + tan θ)

tan + θ tan + θ =
4 4 (1 − tan θ) (1 + tan θ)

π 3π

Hence tan + θ tan + θ = −1.
4 4

1 3π 5 π
24. Find the values of tan(α + β), given that cot α = , α ∈ π, and sec β = − , β ∈ , π .
2 2 3 2
Solution:
s

25 4
Given tan α = 2 and tan β = − 1 = − . Hence,
9 3
4
2 − 2
tan(α + β) = 3 =
8 11
1 +
3
k − 1
25. If θ + φ = α and tan θ = k tan φ, then prove that sin(θ − φ) = sin α.
k + 1

Solution: