Q228 / 270 · NDA 2015 (Phase 2) · Mathematics

If $y=\cot ^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right]$, where $0 \lt x \lt \frac{\pi}{2}$, then $\frac{d y}{d x}$ is equal to

  • Correct answer
    A
    1/2
  • B
    2
  • C
    $\sin x+\cos x$
  • D
    $\sin x-\cos x$

Solution

$\begin{aligned} & y=\cot ^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right] \end{aligned}$ $\begin{aligned} & y=\cot ^{-1}\left[\frac{\sqrt{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^2}+\sqrt{\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)^2}}{\sqrt{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^2}-\sqrt{\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)^2}}\right] \end{aligned}$ $\begin{aligned} & y=\cot ^{-1}\left[\frac{\cos \frac{\mathrm{x}}{2}+\sin \frac{\mathrm{x}}{2}+\cos \frac{\mathrm{x}}{2}-\sin \frac{\mathrm{x}}{2}}{\cos \frac{\mathrm{x}}{2}+\sin \frac{\mathrm{x}}{2}-\cos \frac{\mathrm{x}}{2}+\sin \frac{\mathrm{x}}{2}}\right] \\ & \mathrm{y}=\cot ^{-1}\left[\frac{2 \cos \frac{\mathrm{x}}{2}}{2 \sin \frac{\mathrm{x}}{2}}\right]=\cot ^{-1}\left(\cot \frac{\mathrm{x}}{2}\right)=\frac{\mathrm{x}}{2} \\ & \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2}\end{aligned}$

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

Relevant Questions