$\left( an ^{-1} x ight)^2+\left(\cot ^{-1} x ight)^2=\frac{5 \pi^2}{8} \Rightarrow x=$

Question

$\left(	an ^{-1} xight)^2+\left(\cot ^{-1} xight)^2=\frac{5 \pi^2}{8} \Rightarrow x=$

Aicharya · Accepted Answer

Given equation is $\begin{gathered} \left(	an ^{-1} xight)^2+\left(\cot ^{-1} xight)^2=\frac{5 \pi^2}{8} \ \left(	an ^{-1} x+\cot ^{-1} xight)^2-2 	an ^{-1} x \cdot \cot ^{-1} x=\frac{5 x^2}{8} \ {\left[\because 	an ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}ight]} \end{gathered}$ $\begin{aligned} & \Rightarrow \quad \frac{\pi^2}{4}-2 	an ^{-1} x\left(\pi / 2-	an ^{-1} xight)=\frac{5 \pi^2}{8} \ & \Rightarrow-2 \cdot \frac{\pi 	an ^{-1} x}{2}+2\left(	an ^{-1} xight)^2=\frac{5 \pi^2}{8}-\frac{\pi^2}{4} \ & \Rightarrow \quad\left(	an ^{-1} xight)^2-\frac{\pi 	an ^{-1} x}{2}=\frac{3 \pi^2}{16}\end{aligned}$ $\begin{array}{cc}	herefore \quad & 	an ^{-1} x=\frac{\frac{\pi}{2} \pm \sqrt{\frac{\pi^2}{4}+\frac{3 \pi^2}{4}}}{2} \ \Rightarrow \quad & 	an ^{-1} x=\frac{\frac{\pi}{2} \pm \pi}{2} \ & =\frac{3 \pi}{4},-\frac{\pi}{4}\end{array}$ $\begin{array}{ll}\Rightarrow & x=	an \left(\frac{3 \pi}{4}ight), 	an \left(-\frac{\pi}{4}ight) \ \Rightarrow & x=-1,-1\end{array}$

$\left(\tan ^{-1} x\right)^2+\left(\cot ^{-1} x\right)^2=\frac{5 \pi^2}{8} \Rightarrow x=$

Solution