Question

If $x>y>z>0$, then prove that ${\cot }^{-1}\left(\frac{1+xy}{x-y}\right)+{\cot }^{-1}\left(\frac{1+xy}{y-z}\right)+{\cot }^{-1}\left(\frac{1-zx}{z-x}\right)=\pi$

Answer 1

Since $x>y>z$

$\frac{1+xy}{x-y}>0,\frac{1+xy}{y-z}>0$

but $\frac{1+zx}{z-x}<0$

Hence

${\cot }^{-1}\left(\frac{1+xy}{x-y}\right)+{\cot }^{-1}\left(\frac{1+yz}{y-z}\right)+{\cot }^{-1}\left(\frac{1+zx}{z-x}\right)$

$\Rightarrow {\cot }^{-1}x-{\cot }^{-1}y+{\cot }^{-1}y-{\cot }^{-1}z+\pi -{\cot }^{4}x+{\cot }^{4}z$

$\Rightarrow \pi$