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

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(D) Using the formula $	an^{-1}(A) + 	an^{-1}(B) = 	an^{-1}\left(\frac{A+B}{1-AB}ight)$,we have:$	an^{-1}\left(\frac{\frac{x-1}{x+1} + \frac{2x-

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None of these

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(D) Using the formula $	an^{-1}(A) + 	an^{-1}(B) = 	an^{-1}\left(\frac{A+B}{1-AB}ight)$,we have:$	an^{-1}\left(\frac{\frac{x-1}{x+1} + \frac{2x-1}{2x+1}}{1 - \left(\frac{x-1}{x+1}ight)\left(\frac{2x-1}{2x+1}ight)}ight) = 	an^{-1}\left(\frac{23}{36}ight)$Simplifying the argument:$\frac{(x-1)(2x+1) + (2x-1)(x+1)}{(x+1)(2x+1) - (x-1)(2x-1)} = \frac{23}{36}$$\frac{2x^2 + x - 2x - 1 + 2x^2 + 2x - x - 1}{2x^2 + x + 2x + 1 - (2x^2 - x - 2x + 1)} = \frac{23}{36}$$\frac{4x^2 - 2}{6x} = \frac{23}{36}$$\frac{2x^2 - 1}{3x} = \frac{23}{36}$$24x^2 - 12 = 69x$$24x^2 - 69x - 12 = 0$Dividing by $3$: $8x^2 - 23x - 4 = 0$$(8x + 1)(x - 4) = 0$Thus,$x = 4$ or $x = -\frac{1}{8}$.Since none of the options match,the correct answer is $(d)$.

If $\tan^{-1}\left(\frac{x-1}{x+1}\right) + \tan^{-1}\left(\frac{2x-1}{2x+1}\right) = \tan^{-1}\left(\frac{23}{36}\right)$,then $x =$

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