Question

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

• A. $-1$
• B. $1$
• C. $0$
• D. $2$

Answer 1

The correct option is A $-1$

Let $ta{n}^{-1}x=y$
Therefore
$y^2+(\frac{\pi }{2}-y{)}^2=\frac{5{\pi }^2}{8}$
$2y^2-\frac{2\pi y}{2}+\frac{{\pi }^2}{4}=\frac{5{\pi }^2}{8}$
$2y^2-\pi y-\frac{3{\pi }^2}{8}=0$
$y=\frac{-\pi }{4}$ and $y=\frac{3\pi }{4}$
Hence
$ta{n}^{-1}\left(x\right)=\frac{-\pi }{4}$
$x=-1$ and
$ta{n}^{-1}\left(x\right)=\frac{3\pi }{4}$
$x=-1$
However only $x=-1$ satisfies the above quadratic equation.