MCQ
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
- ✓$1$
- B$2$
- C$3$
- D$4$
✓
Answer
Correct option: A.
$1$
a
$\text { Given, } f(\theta)=\sin \left(\tan ^{-1}(\sin \theta / \sqrt{\cos 2 \theta))}\right.$
$\text { Given, } f(\theta)=\sin \left(\tan ^{-1}(\sin \theta / \sqrt{\cos 2 \theta))}\right.$
$=\sin \left(\sin ^{-1}\left(\frac{\sin \theta}{\sqrt{\sin ^2 \theta+\cos 2 \theta}}\right)\right)$
$=\left(\frac{\sin \theta}{\sqrt{\sin ^2 \theta+\cos ^2 \theta-\sin ^2 \theta}}\right)$
$=\left(\frac{\sin \theta}{\sqrt{\cos ^2 \theta}}\right)$
$=\left(\frac{\sin \theta}{\cos \theta}\right)=\tan \theta$
$\text { Now, } \frac{d}{d(\tan \theta)}(\tan \theta)=1$
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