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