If (tan−1 x)2 + (cot −1x)2=5π28, then the absolute value of x is _______

tan−1⁡x2+cot−1⁡x2=5π28tan−1⁡x+cot−1⁡x2−2tan−1⁡xπ2−tan−1⁡x=5π28π24−2×π2tan−1⁡x+2tan−1⁡x2=5π282tan−1⁡x2−πtan−1⁡x−3π28=0tan−1⁡x=−π4,3π4tan−1⁡x=−π4 ∵tan−1⁡x∈−π2,π2x=−1

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If ${(\tan^{- 1} x)}^{2} + {(c o t^{- 1} x)}^{2} = \frac{5 π^{2}}{8},$ then the absolute value of $x$ is _______

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Detailed Solution

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

$\begin{aligned} {(\tan^{- 1} x + \cot^{- 1} x)}^{2} \\ - 2 \tan^{- 1} x (\frac{π}{2} - \tan^{- 1} x) = \frac{5 π^{2}}{8} \end{aligned}$

$\begin{aligned} \frac{π^{2}}{4} - 2 \times \frac{π}{2} \tan^{- 1} x + 2 {(\tan^{- 1} x)}^{2} = \frac{5 π^{2}}{8} \\ 2 {(\tan^{- 1} x)}^{2} - π \tan^{- 1} x - \frac{3 π^{2}}{8} = 0 \\ \tan^{- 1} x = - \frac{π}{4}, \frac{3 π}{4} \\ \tan^{- 1} x = - \frac{π}{4} (∵ \tan^{- 1} x \in (- \frac{π}{2}, \frac{π}{2})) \\ x = - 1 \end{aligned}$