Question

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

Answer 1

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

Put ${\tan }^{-1}x=c$

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

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

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

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

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

${\tan }^{-1}x=\frac{3\pi }{4},\frac{-\pi }{4}$

$x=\tan \frac{3\pi }{4},\tan \frac{-\pi }{4}$

$x=-1$