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

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