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

\(\begin{aligned} & f(\theta)=\sin \left(\tan ^{-1}\left(\frac{\sin \theta}{\sqrt{\cos 2 \theta}}\right)\right) \\ &=\sin \left(\sin ^{-1}\left(\frac{\sin \theta}{\sqrt{\cos 2 \theta+\sin ^2 \theta}}\right)\right) \\ &=\sin \left(\sin ^{-1}\left(\frac{\left.\cdots \tan ^{-1} x=\sin ^{-1} \frac{x}{\sqrt{1+x^2}}\right]}{\sqrt{\cos ^2 \theta-\sin ^2 \theta+\sin ^2 \theta}}\right)\right) \\ &=\sin \left(\sin ^{-1}\left(\frac{\sin \theta}{\sqrt{\cos ^2 \theta}}\right)\right) \\ &=\sin \left(\sin ^{-1}(\tan \theta)\right) \\ &=\tan \theta \\ & \therefore \quad \frac{d}{d(\tan \theta)}(f(\theta))=1\end{aligned}\)