Q216 / 270 · NDA 2017 (Phase 2) · Mathematics

If $y=\sec ^{-1}\left(\frac{x+1}{x-1}\right)+\sin ^{-1}\left(\frac{x-1}{x+1}\right)$, then $\frac{d y}{d x}$ is equal to

  • Correct answer
    A
    $0$
  • B
    $1$
  • C
    $\frac{\mathrm{x}-1}{\mathrm{x}+1}$
  • D
    $\frac{\mathrm{x}+1}{\mathrm{x}-1}$

Solution

$\begin{aligned} & y=\sec ^{-1}\left(\frac{x+1}{x-1}\right)+\sin ^{-1}\left(\frac{x-1}{x+1}\right) \\ & =\cos ^{-1}\left(\frac{x-1}{x+1}\right)+\sin ^{-1}\left(\frac{x-1}{x+1}\right)\end{aligned}$ $\begin{aligned} & =\frac{\pi}{2}\left(\because \sin ^{-1} \mathrm{x}+\cos ^{-1} \mathrm{x}=\frac{\pi}{2}\right) \\ & \therefore \frac{\mathrm{dy}}{\mathrm{dx}}=0\end{aligned}$

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

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