Question

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

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

Answer 1

The correct option is C $x=0$

${\tan }^{-1}(1+x)+{\tan }^{-1}(1-x)=\frac{\pi }{2}\Rightarrow {\tan }^{-1}(1+x)=\frac{\pi }{2}-{\tan }^{-1}(1-x)\Rightarrow {\tan }^{-1}(1+x)={\cot }^{-1}(1-x)$

$\Rightarrow {\tan }^{-1}(1+x)={\tan }^{-1}\left(\frac{1}{1-x}\right)\Rightarrow 1+x=\frac{1}{1-x}\Rightarrow 1-x^2=1\Rightarrow x=0$