If $ an ^{-1}\left(\frac{1-x}{1+x} ight)=\frac{1}{2} an ^{-1} x$, then $x$ is

Question

If $	an ^{-1}\left(\frac{1-x}{1+x}ight)=\frac{1}{2} 	an ^{-1} x$, then $x$ is

Aicharya · Accepted Answer

$\begin{aligned} & 	an ^{-1}\left(\frac{1-x}{1+x}ight)=\frac{1}{2} 	an ^{-1} x \ & \Rightarrow 	an ^{-1}(1)-	an ^{-1}(x)=\frac{1}{2} 	an ^{-1} x \ & \Rightarrow \frac{\pi}{4}=\frac{3}{2} 	an ^{-1} x \ & \Rightarrow x=	an \left(\frac{\pi}{6}ight)=\frac{1}{\sqrt{3}}\end{aligned}$

If $\tan ^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2} \tan ^{-1} x$, then $x$ is

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