SSC CHSL 20 Dec 2015 Shift 1 Paper

Given,
$\mathit{s}\mathit{i}\mathit{n}\left(90°-\mathit{\theta }\right)+\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }=\sqrt{2}\mathit{c}\mathit{o}\mathit{s}\left(90°-\mathit{\theta }\right)$ ........(1)
We know that, $\sin \left(90°-\mathit{\theta }\right)\cos \mathit{\theta }$ and$\cos \left(90°-\mathit{\theta }\right)\sin \mathit{\theta }$
Now, Equation (1) can be written as,
$\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }+\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }=\sqrt{2}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }$ (or) $2\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }=\sqrt{2}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }$
Squaring on bothsides we get,
$2\mathit{c}\mathit{o}{\mathit{s}}^2\mathit{\theta }=\mathit{s}\mathit{i}{\mathit{n}}^2\mathit{\theta }$
⇒ $2\left(1-\mathit{s}\mathit{i}{\mathit{n}}^2\mathit{\theta }\right)\Rightarrow \mathit{s}\mathit{i}{\mathit{n}}^2\mathit{\theta }$ $\Rightarrow 2-2\mathit{s}\mathit{i}{\mathit{n}}^2\mathit{\theta }=\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }$
$\Rightarrow 2=3\mathit{s}\mathit{i}{\mathit{n}}^2\mathit{\theta }$ $\Rightarrow 2=3\left(\frac{1}{\mathit{c}\mathit{o}\mathit{s}\mathit{e}{\mathit{c}}^2\mathit{\theta }}\right)\Rightarrow \mathit{c}\mathit{o}\mathit{s}\mathit{e}{\mathit{c}}^2\mathit{\theta }=\frac{3}{2}$
⇒ cosecθ = $\sqrt{\frac{3}{2}}$
Hence, option B is the correct answer.