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SSC CHSL 6 Dec 2015 Shift 1 Paper

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Question : 119 of 200

Marks: +1, -0

If

\cos\theta + \sin\theta = m

\sec\theta + \csc\theta = n

then the value of n

(m^2-1)

is equal to :

2n
4mn
mn
2m

Solution:

Given :

\cos\theta + \sin\theta = m

---------(i)

Squaring both sides, we get :

⇒

(\cos\theta + \sin\theta)^2 = m^2

⇒

\cos^2\theta + \sin^2\theta

+ 2\sin\theta\cos\theta = m^2

⇒

1 + 2\sin\theta\cos\theta = m^2

⇒

\sin\theta\cos\theta = \frac{m^2-1}{2}

-------(ii)

Also, it is giventhat :

\sec\theta + \csc\theta = n

⇒

\frac{1}{\cos\theta} + \frac{1}{\sin\theta} = n

⇒

\frac{\sin\theta + \cos\theta}{\sin\theta\cos\theta} = n

Using equations (i) and (ii), ⇒

m = \frac{m^2-1}{2} \times n

⇒

n(m^2-1) = 2m

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