Q63 / 125 · BITSAT 2021 · Mathematics
$\frac{d}{d x}\left(\tan ^{-1} \sqrt{\frac{1+\cos \frac{x}{2}}{1-\cos \frac{x}{2}}}\right)$ is equal to
- Correct answerA$-\frac{1}{4}$
- B$\frac{1}{4}$
- C$-\frac{1}{2}$
- D$\frac{1}{2}$
Solution
We have $\frac{d}{d x}\left(\tan ^{-1} \sqrt{\frac{1+\cos \frac{x}{2}}{1-\cos \frac{x}{2}}}\right)$
$=\frac{d}{d x}\left[\tan ^{-1} \sqrt{\frac{1+\left(2 \cos ^{2} \frac{x}{4}-1\right)}{1-\left(1-2 \sin ^{2} \frac{x}{4}\right)}}\right]$
$=\frac{d}{d x}\left(\tan ^{-1} \sqrt{\left.\frac{2 \cos ^{2} \frac{x}{4}}{2 \sin ^{2} \frac{x}{4}}\right)}\right.$
$=\frac{d}{d x}\left(\tan ^{-1} \sqrt{\cot ^{2} \frac{x}{4}}\right)=\frac{d}{d x}\left(\tan ^{-1} \cot \frac{x}{4}\right)$
$=\frac{d}{d x}\left(\tan ^{-1} \tan \left(\frac{\pi}{2}-\frac{x}{4}\right)\right)=\frac{d}{d x}\left(\frac{\pi}{2}-\frac{x}{4}\right)=0-\frac{1}{4}=-\frac{1}{4}$
Hence, the correct answer is option: 'A'.