Question

The number of solution(s) ${\left({\tan }^{-1}x\right)}^2+{\left({\cot }^{-1}x\right)}^2=\frac{5{\pi }^2}{8}$ is

• A. $1$
• B. $0$
• C. $2$
• D. infinite

Answer 1

The correct option is A $1$

${\left({\tan }^{-1}x\right)}^2+{\left({\cot }^{-1}x\right)}^2=\frac{5{\pi }^2}{8}\Rightarrow {({\tan }^{-1}x+{\cot }^{-1}x)}^2-2{\tan }^{-1}x{\cot }^{-1}x=\frac{5{\pi }^2}{8}\Rightarrow \frac{{\pi }^2}{4}-2{\tan }^{-1}x(\frac{\pi }{2}-{\tan }^{-1}x)=\frac{5{\pi }^2}{8}$
Let ${\tan }^{-1}x=t$
$\Rightarrow -2t(\frac{\pi }{2}-t)=\frac{3{\pi }^2}{8}\Rightarrow 16t^2-8\pi t-3{\pi }^2=0\Rightarrow (4t+\pi )(4t-3\pi )=0\Rightarrow t=-\frac{\pi }{4}(\because t={\tan }^{-1}x\in [-\frac{\pi }{2},\frac{\pi }{2}])\therefore x=-1$
Hence, the number of solution is $1.$