Question

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

• A. $-1$
• B. $0$
• C. $1$
• D. $2$

Answer 1

Answer: (A)

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