Question

Let $f(\theta )=\sin \left({\tan }^{-1}\left(\frac{\sin \theta }{\sqrt{\cos 2\theta }}\right)\right)$, where $-\frac{\pi }{4}<\theta <\frac{\pi }{4}$. Then the value of$\frac{d}{d\left(\tan \theta \right)}\left(f\right(\theta \left)\right)$ is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A $1$

Given $f\left(\theta \right)=\sin \left({\tan }^{-1}\left(\frac{\sin \theta }{\sqrt{\cos 2\theta }}\right)\right)$

Let ${\tan }^{-1}\left(\frac{\sin \theta }{\sqrt{\cos 2\theta }}\right)=y$

$\Rightarrow \tan y=\frac{\sin \theta }{\sqrt{2{\cos }^2\theta -1}}$

Refer to the figure below,
$\Rightarrow \sin y=\frac{\sin \theta }{\cos \theta }$

$f\left(\theta \right)=\sin y=\frac{\sin \theta }{\cos \theta }=\tan \theta$

$\Rightarrow \frac{d}{d\left(\tan \theta \right)}\left(f\left(\theta \right)\right)=1$