Question

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

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is B $1$

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,
$2{\tan }^{-1}x-\frac{\pi }{2}={\cos }^{-1}(2-x)[\because {\tan }^{-1}x+{\cot }^{-1}x=\frac{\pi }{2}]\Rightarrow 2{\tan }^{-1}x=\frac{\pi }{2}+{\cos }^{-1}(2-x)\Rightarrow \pi -{\sin }^{-1}\frac{2x}{(1+x^2)}=\frac{\pi }{2}+{\cos }^{-1}(2-x)[\because {\sin }^{-1}\frac{2x}{(1+x^2)}=\pi -2{\tan }^{-1}x,x\ge 1]\Rightarrow \frac{\pi }{2}-{\sin }^{-1}\frac{2x}{(1+x^2)}={\cos }^{-1}(2-x)$
Applying cosine on both sides
$\Rightarrow \frac{2x}{(1+x^2)}=2-x\Rightarrow x^3-2x^2+3x-2=0\Rightarrow (x-1)(x^2-x+2)=0\Rightarrow x=1$
Hence, the number of solution is $1.$