If tan ^{-1}left(frac{x+1}{x-1}right)+tan ^{- | vedclass

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(C) We use the formula $	an ^{-1}(A) + 	an ^{-1}(B) = 	an ^{-1}\left(\frac{A+B}{1-AB}ight)$.Given $	an ^{-1}\left(\frac{x+1}{x-1}ight)+	an ^{

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$2$

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(C) We use the formula $	an ^{-1}(A) + 	an ^{-1}(B) = 	an ^{-1}\left(\frac{A+B}{1-AB}ight)$.Given $	an ^{-1}\left(\frac{x+1}{x-1}ight)+	an ^{-1}\left(\frac{x-1}{x}ight)=	an ^{-1}(-7)$.Applying the formula,we get $	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)$.Simplifying the expression inside the bracket:$\frac{x(x+1)+(x-1)^2}{x(x-1)-(x+1)(x-1)} = \frac{x^2+x+x^2-2x+1}{x^2-x-(x^2-1)} = \frac{2x^2-x+1}{1-x}$.Equating the arguments,we have $\frac{2x^2-x+1}{1-x} = -7$.$2x^2-x+1 = -7(1-x) = -7+7x$.$2x^2-8x+8 = 0$.$x^2-4x+4 = 0$.$(x-2)^2 = 0$,which gives $x=2$.

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

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