Question

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

• A. $0$
• B. $2$
• C. $1$
• D. $-1$
• E. $2\sqrt{2}$

Answer 1

The correct option is D

$-1$

Explanation for the correct option:

Step1: Solve equation

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

Hence option (4) is correct