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

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(B) We know that $	an ^{-1} x + \cot ^{-1} x = \frac{\pi}{2}$.Let $u = 	an ^{-1} x$. Then $\cot ^{-1} x = \frac{\pi}{2} - u$.The given equation beco

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(B) We know that $	an ^{-1} x + \cot ^{-1} x = \frac{\pi}{2}$.Let $u = 	an ^{-1} x$. Then $\cot ^{-1} x = \frac{\pi}{2} - u$.The given equation becomes $u^2 + (\frac{\pi}{2} - u)^2 = \frac{5 \pi^2}{8}$.Expanding the equation: $u^2 + \frac{\pi^2}{4} - \pi u + u^2 = \frac{5 \pi^2}{8}$.$2u^2 - \pi u + \frac{\pi^2}{4} - \frac{5 \pi^2}{8} = 0$.$2u^2 - \pi u - \frac{3 \pi^2}{8} = 0$.Multiply by $8$: $16u^2 - 8\pi u - 3\pi^2 = 0$.Factoring the quadratic: $(4u - 3\pi)(4u + \pi) = 0$.So,$u = \frac{3\pi}{4}$ or $u = -\frac{\pi}{4}$.Since the range of $	an ^{-1} x$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$,we must have $u = -\frac{\pi}{4}$.Thus,$	an ^{-1} x = -\frac{\pi}{4}$,which implies $x = 	an(-\frac{\pi}{4}) = -1$.Therefore,$x^2 + 1 = (-1)^2 + 1 = 1 + 1 = 2$.

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

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