Question

The sum of all the real solution(s) of ${\tan }^{-1}x-{\cot }^{-1}x={\cos }^{-1}(2-x)$ is

Answer 1

Given : ${\tan }^{-1}x-{\cot }^{-1}x={\cos }^{-1}(2-x)$ Above equality holds good iff
$2-x\in [-1,1]\Rightarrow x\in [1,3]$
Now,
$2{\tan }^{-1}x-\frac{\pi }{2}={\cos }^{-1}(2-x)[\because {\tan }^{-1}x+{\cot }^{-1}x=\frac{\pi }{2}]$
Taking cosine on both sides, we get
$\Rightarrow \cos (2{\tan }^{-1}x-\frac{\pi }{2})=2-x\Rightarrow \sin \left(2{\tan }^{-1}x\right)=2-x\Rightarrow \frac{2\tan \left({\tan }^{-1}x\right)}{1+{\tan }^2\left({\tan }^{-1}x\right)}=2-x\Rightarrow \frac{2x}{(1+x^2)}=2-x\Rightarrow x^3-2x^2+3x-2=0$
By observation, $x=1$ is one root
$\Rightarrow (x-1)(x^2-x+2)=0$
As $x^2-x+2>0\forall x\in R$
$\therefore x=1$
Hence, the sum of all the real solution is $1.$