Question

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

• A. $-1$
• B. $1$
• C. $0$
• D. None of these

Answer 1

The correct option is A $-1$

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

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

$\Rightarrow {\tan }^{-1}x=-\frac{\pi }{4}\Rightarrow x=-1$