Question

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

• A. $1$
• B. $-1$
• C. $0$
• D. not finite

Answer 1

The correct option is B $-1$

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

Since ${\tan }^{-1}x+{\cot }^{-1}x=\frac{\pi }{2},$
So, $({\tan }^{-1}x{)}^2+{(\frac{\pi }{2}-{\tan }^{-1}x)}^2=\frac{5{\pi }^2}{8}$
$\Rightarrow ({\tan }^{-1}x{)}^2+\frac{{\pi }^2}{4}-\pi {\tan }^{-1}x+({\tan }^{-1}x{)}^2=\frac{5{\pi }^2}{8}$
$\Rightarrow 2({\tan }^{-1}x{)}^2-\pi {\tan }^{-1}x+\frac{{\pi }^2}{4}-\frac{5{\pi }^2}{8}=0$
$\Rightarrow 16({\tan }^{-1}x{)}^2-8\pi {\tan }^{-1}x-3{\pi }^2=0$

Above equation is quadratic in ${\tan }^{-1}x$
${\tan }^{-1}x=\frac{8\pi \pm \sqrt{64{\pi }^2+64\cdot 3{\pi }^2}}{2\cdot 16}$
$\Rightarrow {\tan }^{-1}x=\frac{\pi \pm 2\pi }{4}$
$\Rightarrow {\tan }^{-1}x=\frac{3\pi }{4},\frac{-\pi }{4}$
$\Rightarrow {\tan }^{-1}x=\frac{-\pi }{4}$
$\Rightarrow x=-1$