Question

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

Answer 1

Let ${\left({\tan }^{-1}x\right)}^2+{\left({\cot }^{-1}x\right)}^2=\frac{{5\pi }^2}{8}$
$\Rightarrow {({\tan }^{-1}x+{\cot }^{-1}x)}^2-2{\tan }^{-1}x(\left(\frac{\pi }{2}\right)-{\tan }^{-1}x)=\frac{{5\pi }^2}{8}$
$\Rightarrow \left(\frac{{\pi }^2}{4}\right)-2\times \left(\frac{\pi }{2}\right){\tan }^{-1}x+2{\left({\tan }^{-1}x\right)}^2=\frac{{5\pi }^2}{8}$
$\Rightarrow 2{\left({\tan }^{-1}x\right)}^2-\pi {\tan }^{-1}x-\frac{{3\pi }^2}{8}=0$
$\Rightarrow {\tan }^{-1}x=\frac{-\pi }{4},\frac{3\pi }{4}\Rightarrow {\tan }^{-1}x=\frac{-\pi }{4}\Rightarrow x=-1.$