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 $\frac{d}{d(\tan \theta )}(f(\theta ))$ is____

Answer 1

Answer: 1

$\sin \left({\tan }^{-1}\left(\frac{\sin \theta }{\sqrt{\cos 2\theta }}\right)\right)$, where $\theta \in \left(-\frac{\pi }{4},\frac{\pi }{4}\right)$
$\sin \left({\tan }^{-1}\left(\frac{\sin \theta }{\sqrt{2{\cos }^2\theta -1}}\right)\right)$
$=\sin \left({\sin }^{-1}(\tan \theta )\right)=\tan \theta$
$\frac{d(\tan \theta )}{d(\tan \theta )}=1$