If $ an ^{-1} x=\frac{\pi}{4}- an ^{-1}\left(\frac{1}{3} ight)$, then $x$ is

Question

If $	an ^{-1} x=\frac{\pi}{4}-	an ^{-1}\left(\frac{1}{3}ight)$, then $x$ is

Aicharya · Accepted Answer

Given equation $	an ^{-1} x=\frac{\pi}{4}-	an ^{-1}\left(\frac{1}{3}ight)$ $\begin{aligned} &\Rightarrow 	an ^{-1} x=	an ^{-1}(1)-	an ^{-1}\left(\frac{1}{3}ight) \ &\left.\Rightarrow \quad 	an ^{-1} x=	an ^{-1}\left(\frac{1-\frac{\pi}{3}}{1+\frac{1}{3}}ight) 	an ^{-1}(1)ight) \ &\Rightarrow \quad\left[\because 	an ^{-1} x-	an ^{-1} y=	an ^{-1}\left(\frac{x-y}{1+x y}ight)ight] \ &\Rightarrow \quad 	an ^{-1} x=	an ^{-1}\left(\frac{2 / 3}{4 / 3}ight) \ &\Rightarrow \quad x=1 / 2 \end{aligned}$

If $\tan ^{-1} x=\frac{\pi}{4}-\tan ^{-1}\left(\frac{1}{3}\right)$, then $x$ is

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