If $ an ^{-1}\left(\frac{x+1}{x-1} ight)+ an ^{-1}\left(\frac{x-1}{x} ight)= an ^{-1}(-7)$, then $x$ is equal to

Question

If $	an ^{-1}\left(\frac{x+1}{x-1}ight)+	an ^{-1}\left(\frac{x-1}{x}ight)=	an ^{-1}(-7)$, then $x$ is equal to

Aicharya · Accepted Answer

$\begin{aligned} & 	an ^{-1}\left(\frac{x+1}{x-1}ight)+	an ^{-1}\left(\frac{x-1}{x}ight)=	an ^{-1}(-7) \ & \Rightarrow 	an ^{-1}\left[\frac{\frac{x+1}{x-1}+\frac{x-1}{x}}{1-\left(\frac{x+1}{x-1}ight)\left(\frac{x-1}{x}ight)}ight]=	an ^{-1}(-7)\end{aligned}$ $\begin{aligned} & \Rightarrow 	an ^{-1}\left[\frac{x(x+1)+(x-1)(x-1)}{x(x-1)-(x+1)(x-1)}ight]=	an ^{-1}(-7) \ & \Rightarrow 	an ^{-1}\left(\frac{x^2+x+x^2-2 x+1}{x^2-x-x^2+1}ight)=	an ^{-1}(-7) \ & \Rightarrow 	an ^{-1}\left(\frac{2 x^2-x+1}{1-x}ight)=	an ^{-1}(-7)\end{aligned}$ $\begin{aligned} & \Rightarrow \frac{2 x^2-x+1}{1-x}=-7 \ & \Rightarrow 2 x^2-8 x+8=0 \ & \Rightarrow x^2-4 x+4=0 \ & \Rightarrow(x-2)^2=0 \ & \Rightarrow x=2\end{aligned}$

If $\tan ^{-1}\left(\frac{x+1}{x-1}\right)+\tan ^{-1}\left(\frac{x-1}{x}\right)=\tan ^{-1}(-7)$, then $x$ is…

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