Question

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

Answer 1

Given that

$(ta{n}^{-1}x{)}^2+(co{t}^{-1}x{)}^2=\frac{5{\pi }^2}{8}$

We know that,

$ta{n}^{-1}x+co{t}^{-1}x=\frac{\pi }{2}$ .....$\left(1\right)$

Squaring on both sides

$\Rightarrow (ta{n}^{-1}x+co{t}^{-1}x{)}^2=(\frac{\pi }{2}{)}^2$

$\Rightarrow (ta{n}^{-1}x{)}^2+(co{t}^{-1}x{)}^2+2\left(ta{n}^{-1}x\right)\left(co{t}^{-1}x\right)=\frac{{\pi }^2}{4}$ ....$\left(2\right)$

from the given information,

$(ta{n}^{-1}x{)}^2+(co{t}^{-1}x{)}^2=\frac{5{\pi }^2}{8}$

Substituting this in $\left(2\right)$, we get

$\Rightarrow \frac{5{\pi }^2}{8}+2\left(ta{n}^{-1}x\right)\left(co{t}^{-1}x\right)=\frac{{\pi }^2}{4}$

$\Rightarrow 2\left(ta{n}^{-1}x\right)\left(co{t}^{-1}x\right)=\frac{-3{\pi }^2}{8}$ ....$\left(3\right)$

We know that $(a-b{)}^2=(a+b{)}^2-4ab$

Using this formula, we obtain

$(ta{n}^{-1}x-co{t}^{-1}x{)}^2=(ta{n}^{-1}x+co{t}^{-1}x{)}^2-4\left(ta{n}^{-1}x\right)\left(co{t}^{-1}x\right)$

$\Rightarrow (ta{n}^{-1}x-co{t}^{-1}x{)}^2=\frac{{\pi }^2}{4}-2(\frac{-3{\pi }^2}{8})$ $\left(from\right(2),(3\left)\right)$

$\Rightarrow (ta{n}^{-1}x-co{t}^{-1}x{)}^2=\frac{{\pi }^2}{4}+\frac{3{\pi }^2}{4}$

$(ta{n}^{-1}x-co{t}^{-1}x{)}^2={\pi }^2$

Taking square root,

$\Rightarrow ta{n}^{-1}x-co{t}^{-1}x=\pi$ .....$\left(4\right)$

On solving $\left(1\right)$ and $\left(4\right)$, we get

$\Rightarrow 2ta{n}^{-1}x=\frac{3\pi }{2}$

$\Rightarrow ta{n}^{-1}x=\frac{3\pi }{4}$

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

$\Rightarrow x=-1$