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)

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