If Tan-1 x2+Cot-1 x2=5π28 then x=

Tan-1 x+Cot-1 x2-2Tan-1 xπ2-Tan-1 x=5π28 ⇒2Tan-1 x2-πTan-1 x-3π28=0⇒x=-1

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If ${({Tan}^{- 1} x)}^{2} + {({Cot}^{- 1} x)}^{2} = \frac{5 π^{2}}{8}$ then x=

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answer is -1.

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Detailed Solution

${({Tan}^{- 1} x + {Cot}^{- 1} x)}^{2} - 2 {Tan}^{- 1} x (\frac{π}{2} - {Tan}^{- 1} x) = \frac{5 π^{2}}{8}$ $\Rightarrow 2 {({Tan}^{- 1} x)}^{2} - {πTan}^{- 1} x - \frac{3 π^{2}}{8} = 0 \Rightarrow x = - 1$

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