JEE Advanced · Mathematics · 21. ITF
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
- A 1
- B 5
- C 9
- D 10
Answer & Solution
Correct Answer
(A) 1
Step-by-step Solution
Detailed explanation
\(f(\theta)=\sin \left(\tan ^{-1} \frac{\sin \theta}{\sqrt{\cos 2 \theta}}\right),-\frac{\pi}{4} < \theta < \frac{\pi}{4}\)
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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