Question

If $x>y>z>0$, then find the value of ${\cot }^{-1}\frac{xy+1}{x-y}+{\cot }^{-1}\frac{yz+1}{y-z}+{\cot }^{-1}\frac{zx+1}{z-x}$

• A. $-\pi /2$
• B. $+\pi /2$
• C. $-\pi$
• D. $+\pi$

Answer 1

The correct option is D $+\pi$

Given, ${\cot }^{-1}\frac{xy+1}{x-y}+{\cot }^{-1}\frac{yz+1}{y-z}+{\cot }^{-1}\frac{zx+1}{z-x}$

$={\tan }^{-1}\frac{x-y}{1+xy}+{\tan }^{-1}\frac{y-z}{1+yz}+\pi +{\tan }^{-1}\frac{z-x}{1+zx}$ $[\because {\tan }^{-1}\left(\frac{1}{x}\right)=\{\begin{array}{c}{\cot }^{-1}x,forx>0\\ -\pi +{\cot }^{-1}x,forx<0\end{array}]$

$={\tan }^{-1}x-{\tan }^{-1}y+{\tan }^{-1}y-{\tan }^{-1}z+\pi +{\tan }^{-1}z-{\tan }^{-1}x=\pi$