Question

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

• A. $0$
• B. $-1$
• C. $-2$
• D. $-3$

Answer 1

The correct option is B $-1$

$({\tan }^{-1}x{)}^2+({\cot }^{-1}x{)}^2=\frac{5{\pi }^2}{8}$
$\Rightarrow ({\tan }^{-1}x+{\cot }^{-1}x{)}^2-2{\tan }^{-1}x{\cot }^{-1}x=\frac{5{\pi }^2}{8}$
$\Rightarrow \frac{{\pi }^2}{4}-2{\tan }^{-1}x{\cot }^{-1}x=\frac{5{\pi }^2}{8}$
$\Rightarrow {\tan }^{-1}x{\cot }^{-1}x=\frac{-3{\pi }^2}{16}\cdots \left(1\right)$
$\text{Let}{\tan }^{-1}x=t$
$\text{Then, equation (1) becomes}$
$t(\frac{\pi }{2}-t)=\frac{-3{\pi }^2}{16}$
$\Rightarrow t=\frac{-\pi }{4},\frac{3\pi }{4}$
$\therefore x=\tan t=-1$