Q92 / 150 · MHT CET 2020 (14 Oct Shift 1) · Mathematics

If $y=\tan ^{-1}\left[\sqrt{\frac{1+\cos \frac{x}{2}}{1-\cos \frac{x}{2}}}\right]$, then $\frac{d y}{d x}=$

  • A
    $\frac{-1}{3}$
  • Correct answer
    B
    $\frac{-1}{4}$
  • C
    $\frac{1}{3}$
  • D
    $\frac{1}{4}$

Solution

Given $y=\tan ^{-1}\left[\sqrt{\frac{1+\cos \frac{x}{2}}{1-\cos \frac{x}{2}}}\right]$ $$ =\tan ^{-1} \sqrt{\frac{2 \cos ^{2} \frac{x}{4}}{2 \sin ^{2} \frac{x}{4}}}=\tan ^{-1}\left(\cot \frac{x}{4}\right)=\tan ^{-1}\left[\tan \left(\frac{\pi}{2}-\frac{x}{4}\right)\right] $$ $\therefore y=-\frac{x}{4} \Rightarrow \frac{d y}{d x}=\frac{-1}{4}$

Hence, the correct answer is option: 'B'.

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