Question

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

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

Answer 1

Answer: (A)

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