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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
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$f(\theta)=\sin \left(\tan ^{-1} \frac{\sin \theta}{\sqrt{\cos 2 \theta}}\right),-\frac{\pi}{4} < \theta < \frac{\pi}{4}$
Let $\tan ^{-1} \frac{\sin \theta}{\sqrt{\cos 2 \theta}}=\phi$
$$
\Rightarrow \quad \tan \phi=\frac{\sin \theta}{\sqrt{\cos 2 \theta}}
$$
$$
\begin{aligned}
\therefore \quad \sin \phi & =\frac{\sin \theta}{\sqrt{\sin ^2 \theta+\cos 2 \theta}} \\
& =\frac{\sin \theta}{\sqrt{1-\sin ^2 \theta}}=\frac{\sin \theta}{\cos \theta}=\tan \theta
\end{aligned}
$$
$$
\begin{array}{ll}
\therefore & f(\theta)=\sin \phi=\tan \theta \\
\Rightarrow & \frac{d f(\theta)}{d(\tan \theta)}=1
\end{array}
$$
Let $\tan ^{-1} \frac{\sin \theta}{\sqrt{\cos 2 \theta}}=\phi$
$$
\Rightarrow \quad \tan \phi=\frac{\sin \theta}{\sqrt{\cos 2 \theta}}
$$
$$
\begin{aligned}
\therefore \quad \sin \phi & =\frac{\sin \theta}{\sqrt{\sin ^2 \theta+\cos 2 \theta}} \\
& =\frac{\sin \theta}{\sqrt{1-\sin ^2 \theta}}=\frac{\sin \theta}{\cos \theta}=\tan \theta
\end{aligned}
$$
$$
\begin{array}{ll}
\therefore & f(\theta)=\sin \phi=\tan \theta \\
\Rightarrow & \frac{d f(\theta)}{d(\tan \theta)}=1
\end{array}
$$
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