If (tan^{-1} x)^2 + (cot^{-1} x)^2 = frac{5pi | vedclass

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Question

(A) We know that $\cot^{-1} x = \frac{\pi}{2} - 	an^{-1} x$.Substituting this into the given equation:$(	an^{-1} x)^2 + (\frac{\pi}{2} - 	an^{-1} x

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(A) We know that $\cot^{-1} x = \frac{\pi}{2} - 	an^{-1} x$.Substituting this into the given equation:$(	an^{-1} x)^2 + (\frac{\pi}{2} - 	an^{-1} x)^2 = \frac{5\pi^2}{8}$.Let $t = 	an^{-1} x$. Then the equation becomes:$t^2 + (\frac{\pi}{2} - t)^2 = \frac{5\pi^2}{8}$.Expanding the square:$t^2 + \frac{\pi^2}{4} - \pi t + t^2 = \frac{5\pi^2}{8}$.$2t^2 - \pi t + \frac{\pi^2}{4} - \frac{5\pi^2}{8} = 0$.$2t^2 - \pi t - \frac{3\pi^2}{8} = 0$.Multiplying by $8$ to clear the denominator:$16t^2 - 8\pi t - 3\pi^2 = 0$.Factoring the quadratic equation:$(4t - 3\pi)(4t + \pi) = 0$.So,$t = \frac{3\pi}{4}$ or $t = -\frac{\pi}{4}$.Since the range of $	an^{-1} x$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$,we must have $	an^{-1} x = -\frac{\pi}{4}$.Therefore,$x = 	an(-\frac{\pi}{4}) = -1$.

If $(\tan^{-1} x)^2 + (\cot^{-1} x)^2 = \frac{5\pi^2}{8}$,then $x$ =

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