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)

${\left(ta{n}^{-1}x\right)}^2+\left(co{t}^{-1}{x}^2\right)=\frac{5{\pi }^2}{8}$
$\Rightarrow {\left(ta{n}^{-1}x+co{t}^{-1}x\right)}^2-2ta{n}^{-1}x$
$\left(\frac{\pi }{2}-ta{n}^{-1}x\right)=\frac{5{\pi }^2}{8}$
$\Rightarrow {\left(\frac{\pi }{2}\right)}^2-2\left(\frac{\pi }{2}\right)ta{n}^{-1}x+2{\left(ta{n}^{-1}x\right)}^2$
$=\frac{5{\pi }^2}{8}$
$\Rightarrow 2{\left(ta{n}^{-1}x\right)}^2-\pi ta{n}^{-1}x-\frac{3{\pi }^2}{8}=0$
$\Rightarrow ta{n}^{-1}x=-\frac{\pi }{4},\frac{3\pi }{4}$
$\Rightarrow ta{n}^{-1}x=-\pi /4$
$\Rightarrow x=-1$