If $ an ^{-1}\left(\frac{1-x}{1+x} ight)-\frac{1}{2} an ^{-1} x=0$, for $x0$, then $x=$

Question

If $	an ^{-1}\left(\frac{1-x}{1+x}ight)-\frac{1}{2} 	an ^{-1} x=0$, for $x>0$, then $x=$

Aicharya · Accepted Answer

Here $	an ^{-1}\left(\frac{1-x}{1+x}ight)=\frac{1}{2} 	an ^{-1} x \Rightarrow 	an ^{-1}\left[\frac{1-x}{1+(1)(x)}ight]=\frac{1}{2} 	an ^{-1} x$ $	herefore 	an ^{-1}(1)-	an ^{-1} x=\frac{1}{2} 	an ^{-1} x$ $\frac{\pi}{4}=\frac{3}{2} 	an ^{-1} x \Rightarrow 	an ^{-1} x=\frac{\pi}{4} 	imes \frac{2}{3} \Rightarrow 	an ^{-1} x=\frac{\pi}{6}$ $	herefore x=	an \frac{\pi}{6}=\frac{1}{\sqrt{3}}$

If $\tan ^{-1}\left(\frac{1-x}{1+x}\right)-\frac{1}{2} \tan ^{-1} x=0$, for $x>0$, then $x=$

Solution