SSC CHSL 6 Dec 2015 Shift 1 Paper

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Question : 119 of 200

Marks: +1, -0

If

Cosθ+Sinθ=m

and

Secθ+Cosecθ=n

then the value of n

(m^2-1)

is equal to :

2n
4mn
mn
2m

Solution:

Given :

Cosθ+Sinθ=m

---------(i)
Squaring both sides, we get :
⇒

(cos θ + sin θ) ^2 = (m) ^2

⇒

cos ^2 θ + sin^ 2 θ

+ 2sin θ.cos θ = m^2

⇒

1 + 2sin θ.cos θ = m^2

⇒

sinθ.cosθ={m^2-1}/2

-------(ii)
Also, it is giventhat :

sec θ + cosec θ = n

⇒

1/{cos θ} + 1 /{sin θ }=n

⇒

{sin θ+cos θ}/{sin θ.cos θ }=n

Using equations (i) and (ii), ⇒

m = {m^2 −1}/2 × n

⇒

n(m^2 − 1) = 2m

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