Question

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

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

Answer 1

Answer: (D)

Given, ${({\tan }^{-1}x)}^2+{({\cot }^{-1}x)}^2=\frac{5{\pi }^2}{8}$
$\therefore$ ${({\tan }^{-1}x+{\cot }^{-1}x)}^2-2{\tan }^{-1}x\left(\frac{\pi }{2}-{\tan }^{-1}x\right)$
$=\frac{5{\pi }^2}{8}$
$\Rightarrow$ $\frac{{\pi }^2}{4}-2\times \frac{\pi }{2}{\tan }^{-1}x+2{({\tan }^{-1}x)}^2=\frac{5{\pi }^2}{8}$
$\Rightarrow$ $2{({\tan }^{-1}x)}^2-\pi {\tan }^{-1}x-\frac{3{\pi }^2}{8}=0$
$\Rightarrow$ ${\tan }^{-1}x=-\frac{\pi }{4},\frac{3\pi }{4}$
Now, we take ${\tan }^{-1}x=-\frac{\pi }{4}\Rightarrow x=-1$