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

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(B) Given: $(	an ^{-1} x)^2+(\cot ^{-1} x)^2=\frac{5 \pi^2}{8}$ We know that $	an ^{-1} x + \cot ^{-1} x = \frac{\pi}{2}$. Let $u = 	an ^{-1} x$. T

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(B) Given: $(	an ^{-1} x)^2+(\cot ^{-1} x)^2=\frac{5 \pi^2}{8}$ 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 equation becomes $u^2 + (\frac{\pi}{2} - u)^2 = \frac{5 \pi^2}{8}$. Expanding the terms: $u^2 + \frac{\pi^2}{4} - \pi u + u^2 = \frac{5 \pi^2}{8}$. $2u^2 - \pi u + \frac{\pi^2}{4} - \frac{5 \pi^2}{8} = 0$. $2u^2 - \pi u - \frac{3 \pi^2}{8} = 0$. Multiply by $8$: $16u^2 - 8\pi u - 3\pi^2 = 0$. Factoring the quadratic: $(4u + \pi)(4u - 3\pi) = 0$. So,$u = -\frac{\pi}{4}$ or $u = \frac{3\pi}{4}$. Since the range of $	an ^{-1} x$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$,we must have $u = -\frac{\pi}{4}$. Therefore,$	an ^{-1} x = -\frac{\pi}{4} \Rightarrow x = 	an(-\frac{\pi}{4}) = -1$.

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

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