If {({tan ^{ - 1}}x)^2} + {({cot ^{ - 1}}x)^2 | vedclass

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(A) We are given the equation: ${({	an ^{ - 1}}x)^2} + {({\cot ^{ - 1}}x)^2} = \frac{{5{\pi ^2}}}{8}$.Using the identity ${	an ^{ - 1}}x + {\cot ^{

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(A) We are given the equation: ${({	an ^{ - 1}}x)^2} + {({\cot ^{ - 1}}x)^2} = \frac{{5{\pi ^2}}}{8}$.Using the identity ${	an ^{ - 1}}x + {\cot ^{ - 1}}x = \frac{\pi }{2}$,we can write ${\cot ^{ - 1}}x = \frac{\pi }{2} - {	an ^{ - 1}}x$.Substituting this into the equation:${({	an ^{ - 1}}x)^2} + {(\frac{\pi }{2} - {	an ^{ - 1}}x)^2} = \frac{{5{\pi ^2}}}{8}$.Let $u = {	an ^{ - 1}}x$. Then the equation becomes:$u^2 + (\frac{\pi }{2} - u)^2 = \frac{{5{\pi ^2}}}{8}$.Expanding the square:$u^2 + \frac{{{\pi ^2}}}{4} - \pi u + u^2 = \frac{{5{\pi ^2}}}{8}$.$2u^2 - \pi u + \frac{{{\pi ^2}}}{4} - \frac{{5{\pi ^2}}}{8} = 0$.$2u^2 - \pi u - \frac{{3{\pi ^2}}}{8} = 0$.Multiplying by $8$ to clear the denominator:$16u^2 - 8\pi u - 3{\pi ^2} = 0$.Solving the quadratic equation for $u$ using the quadratic formula $u = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$:$u = \frac{{8\pi \pm \sqrt {{{( - 8\pi )}^2} - 4(16)( - 3{\pi ^2})} }}{{32}} = \frac{{8\pi \pm \sqrt {64{\pi ^2} + 192{\pi ^2}} }}{{32}} = \frac{{8\pi \pm \sqrt {256{\pi ^2}} }}{{32}} = \frac{{8\pi \pm 16\pi }}{{32}}$.So,$u = \frac{{24\pi }}{{32}} = \frac{{3\pi }}{4}$ or $u = \frac{{ - 8\pi }}{{32}} = - \frac{\pi }{4}$.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 $x$ equals

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