Question

Find the value of x. ${\left({\tan }^{-1}x\right)}^2+{\left({\cot }^{-1}x\right)}^2=\frac{5{\pi }^2}{8}$

Answer 1

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

Let ${\tan }^{-1}x=k$

$2k^2-\pi k+\frac{{\pi }^2}{4}-\frac{5{\pi }^2}{8}=0$

$16k^2-8\pi k-3{\pi }^2=0$

$16k^2-12\pi k+4\pi k-3{\pi }^2=0$

$(4k-3\pi )(4k+\pi )=0$

$k=\frac{3\pi }{4},k=\frac{-\pi }{4}$

$x=\tan \left(\frac{3\pi }{4}\right)$, $x=\tan \left(\frac{-\pi }{4}\right)$

$x=-1$