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

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(A) Given equation is $(	an ^{-1} x)^2+(\cot ^{-1} x)^2=\frac{5 \pi^2}{8}$.Using the identity $	an ^{-1} x + \cot ^{-1} x = \frac{\pi}{2}$,we can wr

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(A) Given equation is $(	an ^{-1} x)^2+(\cot ^{-1} x)^2=\frac{5 \pi^2}{8}$.Using the identity $	an ^{-1} x + \cot ^{-1} x = \frac{\pi}{2}$,we can write $(	an ^{-1} x)^2 + (\cot ^{-1} x)^2 = (	an ^{-1} x + \cot ^{-1} x)^2 - 2 	an ^{-1} x \cot ^{-1} x$.Substituting the identity: $(\frac{\pi}{2})^2 - 2 	an ^{-1} x (\frac{\pi}{2} - 	an ^{-1} x) = \frac{5 \pi^2}{8}$.$\frac{\pi^2}{4} - \pi 	an ^{-1} x + 2(	an ^{-1} x)^2 = \frac{5 \pi^2}{8}$.$2(	an ^{-1} x)^2 - \pi 	an ^{-1} x + \frac{\pi^2}{4} - \frac{5 \pi^2}{8} = 0$.$2(	an ^{-1} x)^2 - \pi 	an ^{-1} x - \frac{3 \pi^2}{8} = 0$.Let $u = 	an ^{-1} x$. Then $2u^2 - \pi u - \frac{3 \pi^2}{8} = 0$.Using the quadratic formula $u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$u = \frac{\pi \pm \sqrt{\pi^2 - 4(2)(-\frac{3 \pi^2}{8})}}{4} = \frac{\pi \pm \sqrt{\pi^2 + 3 \pi^2}}{4} = \frac{\pi \pm 2 \pi}{4}$.So,$u = \frac{3 \pi}{4}$ or $u = -\frac{\pi}{4}$.Since $x = 	an u$,we have $x = 	an(\frac{3 \pi}{4}) = -1$ or $x = 	an(-\frac{\pi}{4}) = -1$.Therefore,$x = -1$.

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

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