The solution of the equation $\tan ^{-1}(1+x)+\tan ^{-1}(1-x)=\frac{\pi}{2}$ is
The solution of the equation $\tan ^{-1}(1+x)+\tan ^{-1}(1-x)=\frac{\pi}{2}$ is
- $x=1$
- $x=0$
- $x=-1$
- $x=\pi$
Solution
$\begin{aligned} & \tan ^{-1}(1+x)+\tan ^{-1}(1-x)=\frac{\pi}{2} \\ & \Rightarrow \tan ^{-1}(1+x)=\frac{\pi}{2}-\tan ^{-1}(1-x) \\ & \Rightarrow \tan ^{-1}(1+x)=\cot ^{-1}(1-x) \\ & \Rightarrow \tan ^{-1}(1+x)=\tan ^{-1}\left(\frac{1}{1-x}\right) \\ & \Rightarrow 1+x=\frac{1}{1-x} \Rightarrow 1-x^2=1 \Rightarrow x=0\end{aligned}$
Asked in: MHT CET 2023 (10 May Shift 1)
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