Q76 / 150 · MHT CET 2023 (10 May Shift 1) · Mathematics

The solution of the equation $\tan ^{-1}(1+x)+\tan ^{-1}(1-x)=\frac{\pi}{2}$ is

  • A
    $x=1$
  • Correct answer
    B
    $x=0$
  • C
    $x=-1$
  • D
    $x=\pi$

Solution

$\begin{aligned} & \tan ^{-1}(1+x)+\tan ^{-1}(1-x)=\frac{\pi}{2} \\ & \Rightarrow \tan ^{-1}(1+x)=\frac{\pi}{2}-\tan ^{-1}(1-x) \\ & \Rightarrow \tan ^{-1}(1+x)=\cot ^{-1}(1-x) \\ & \Rightarrow \tan ^{-1}(1+x)=\tan ^{-1}\left(\frac{1}{1-x}\right) \\ & \Rightarrow 1+x=\frac{1}{1-x} \Rightarrow 1-x^2=1 \Rightarrow x=0\end{aligned}$

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

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