Q156 / 270 · NDA 2015 (Phase 1) · Mathematics

The equation $\tan ^{-1}(1+x)+\tan ^{-1}(1-x)=\frac{\pi}{2}$ is satisfied by

  • A
    $x=1$
  • B
    $x=-1$
  • Correct answer
    C
    $\mathrm{x}=0$
  • D
    $x=\frac{1}{2}$

Solution

$\begin{aligned} & \tan ^{-1}(1+x)+\tan ^{-1}(1-x)=\frac{\pi}{2} \\ & \tan ^{-1}\left[\frac{(1+x)+(1-x)}{1-(1+x)(1-x)}\right]=\frac{\pi}{2}\end{aligned}$ $\begin{array}{ll} \Rightarrow & \frac{1+x+1-x}{1-(1+x)(1-x)}=\tan \frac{\pi}{2} \\ \Rightarrow & \frac{2}{1-(1+x)(1-x)}=\frac{1}{0} \\ \Rightarrow & 1-(1+x)(1-x)=0 \\ \Rightarrow & (1+x)(1-x)=1 \\ & 1-x^2=1 \\ & x^2=0 \\ & x=0 \end{array}$ $\therefore \quad$ Option (C) is correct.

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

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