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

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(A) We know that $	an^{-1} x + \cot^{-1} x = \frac{\pi}{2}$.Let $a = 	an^{-1} x$ and $b = \cot^{-1} x$. Then $a + b = \frac{\pi}{2}$, so $b = \frac{

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(A) We know that $	an^{-1} x + \cot^{-1} x = \frac{\pi}{2}$.Let $a = 	an^{-1} x$ and $b = \cot^{-1} x$. Then $a + b = \frac{\pi}{2}$, so $b = \frac{\pi}{2} - a$.The given equation is $a^2 + b^2 = \frac{5\pi^2}{8}$.Substituting $b$, we get $a^2 + (\frac{\pi}{2} - a)^2 = \frac{5\pi^2}{8}$.$a^2 + \frac{\pi^2}{4} - \pi a + a^2 = \frac{5\pi^2}{8}$.$2a^2 - \pi a + \frac{2\pi^2}{8} - \frac{5\pi^2}{8} = 0$.$2a^2 - \pi a - \frac{3\pi^2}{8} = 0$.Multiply by $8$: $16a^2 - 8\pi a - 3\pi^2 = 0$.Using the quadratic formula $a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, we get $a = \frac{8\pi \pm \sqrt{64\pi^2 - 4(16)(-3\pi^2)}}{32} = \frac{8\pi \pm \sqrt{64\pi^2 + 192\pi^2}}{32} = \frac{8\pi \pm \sqrt{256\pi^2}}{32} = \frac{8\pi \pm 16\pi}{32}$.Case $1$: $a = \frac{24\pi}{32} = \frac{3\pi}{4}$. Since the range of $	an^{-1} x$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$, this is invalid.Case $2$: $a = \frac{-8\pi}{32} = -\frac{\pi}{4}$.Thus, $	an^{-1} x = -\frac{\pi}{4}$, which implies $x = 	an(-\frac{\pi}{4}) = -1$.

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

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