Question

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

• A. -1
• B. 1
• C. 0
• D. $\pi \sqrt{\frac{5}{8}}$

Answer 1

Answer: (A)

Given equation is
$\left(\tan ^{-1} x\right)^{2}+\left(\cot ^{-1} x\right)^{2}=\frac{5 \pi^{2}}{8} $
$\left(\tan ^{-1} x+\cot ^{-1} x\right)^{2}-2 \tan ^{-1} x \cdot \cot ^{-1} x=\frac{5 x^{2}}{8} \,\,\, \left[\because \tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}\right] $
$ \Rightarrow \frac{\pi^{2}}{4}-2 \tan ^{-1} x\left(\pi / 2-\tan ^{-1} x\right)=\frac{5 \pi^{2}}{8} $
$\Rightarrow -2 \cdot \frac{\pi \tan ^{-1} x}{2}+2\left(\tan ^{-1} x\right)^{2}=\frac{5 \pi^{2}}{8}-\frac{\pi^{2}}{4} $
$\Rightarrow \left(\tan ^{-1} x\right)^{2}-\frac{\pi \tan ^{-1} x}{2}=\frac{3 \pi^{2}}{16} $
$\therefore \tan ^{-1} x=\frac{\frac{\pi}{2} \pm \sqrt{\frac{\pi^{2}}{4}+\frac{3 \pi^{2}}{4}}}{2}$
$\Rightarrow \tan ^{-1} x=\frac{\frac{\pi}{2} \pm \pi}{2}$
$=\frac{3 \pi}{4},-\frac{\pi}{4}$
$\Rightarrow x=\tan \left(\frac{3 \pi}{4}\right), \tan \left(-\frac{\pi}{4}\right)$
$\Rightarrow x=-1,-1$