Question

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

Answer 1

$\begin{array}{c}{\left({\tan }^{-1}x\right)}^2+{\left({\cot }^{-1}x\right)}^2=\frac{5{\pi }^2}{2}\\ {\left({\tan }^{-1}x\right)}^2{\left(\frac{\pi }{2}-{\tan }^{-1}x\right)}^2=\frac{5\pi }{8}\\ {\left({\tan }^{-1}x\right)}^2+\frac{{\pi }^2}{4}+{\left({\tan }^{-1}x\right)}^2+\pi {\tan }^{-1}x=\frac{5{\pi }^2}{8}\\ 2{\left({\tan }^{-1}x\right)}^2+\frac{\pi }{4}-\frac{5{\pi }^2}{8}-\pi {\tan }^{-1}x=0\\ 16{\left({\tan }^{-1}x\right)}^2-8\pi {\tan }^{-1}x-3{\pi }^2=0\\ \left(4ta{b}^{-1}x-3\pi \right)\left(4{\tan }^{-1}x+\pi \right)=0\\ {\tan }^{-1}x=\frac{3\pi }{4},{\tan }^{-1}x=-\frac{\pi }{4}\\ x=\tan \frac{3\pi }{4},x=\tan \left(\frac{-\pi }{4}\right)\\ x=-1,x=-1\\ Hence,x=-1.\end{array}$