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

Question

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

Aicharya · Accepted Answer

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

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

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