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(tan ^{-1 _x} ) ² + (cot- ¹ _^x ) ² = 5 pi ² / 8

find value of x

$Given (\tan^{- 1} x)^{2} + (\cot^{- 1} x)^{2} = \frac{5 π^{2}}{8} \Rightarrow (\tan^{- 1} x + \cot^{- 1} x)^{2} + 2 \tan^{- 1} x \cot^{- 1} x = \frac{5 π^{2}}{8} \Rightarrow (\frac{π}{2})^{2} + 2 \tan^{- 1} x (π - \tan^{- 1} x) = \frac{5 π^{2}}{8} \Rightarrow \frac{π^{2}}{4} + 2 π \tan^{- 1} x - 2 (\tan^{- 1} x)^{2} = \frac{5 π^{2}}{8} \Rightarrow 2 π \tan^{- 1} x - 2 (\tan^{- 1} x)^{2} + \frac{π^{2}}{4} - \frac{5 π^{2}}{8} = 0 \Rightarrow 2 (\tan^{- 1} x)^{2} - 2 π \tan^{- 1} x - \frac{π^{2}}{4} + \frac{5 π^{2}}{8} = 0 \Rightarrow 2 (\tan^{- 1} x)^{2} - 2 π \tan^{- 1} x + \frac{5 π^{2} - 2 π^{2}}{8} = 0 \Rightarrow 2 (\tan^{- 1} x)^{2} - 2 π \tan^{- 1} x + \frac{3 π^{2}}{8} = 0 solving the quadratic equation \Rightarrow \tan^{- 1} x = \frac{2 π \pm \sqrt{4 π^{2} - 4 \times 2 \times \frac{3 π^{2}}{8}}}{2 \times 2} \Rightarrow \tan^{- 1} x = \frac{2 π \pm \sqrt{4 π^{2} - 3 π^{2}}}{4} \Rightarrow \tan^{- 1} x = \frac{2 π \pm π}{4} \Rightarrow \tan^{- 1} x = \frac{2 π + π}{4} or \tan^{- 1} x = \frac{2 π - π}{4} \Rightarrow \tan^{- 1} x = \frac{3 π}{4} {or tan}^{- 1} x = \frac{π}{4} \Rightarrow x = \tan \frac{3 π}{4} or x=tan \frac{π}{4} \Rightarrow x = 0, 1$

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