Question

If ${\cot }^{-1}\frac{1}{x}+{\cot }^{-1}\frac{1}{y}+{\cot }^{-1}\frac{1}{z}=\frac{\pi }{2}$, then $xy+yz+zx=$

Answer 1

${\cot }^{-1}\frac{1}{x}+{\cot }^{-1}\frac{1}{y}+{\cot }^{-1}\frac{1}{z}=\frac{\pi }{2}$

${\cot }^{-1}\frac{1}{x}+{\cot }^{-1}\frac{1}{y}=\frac{\pi }{2}-{\cot }^{-1}\frac{1}{z}$

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

${\tan }^{-1}\left(\frac{x+y}{1-xy}\right)={\tan }^{-1}\frac{1}{z}$

$\Rightarrow \left(\frac{x+y}{1-xy}\right)=\frac{1}{z}$

$xz+yz=1-xy$

$xy+xz+yz=1$