Question

If ${\sec }^{-1}\left(\sqrt{1+x^2}\right)+co{\sec }^{-1}\left(\frac{\sqrt{1+y^2}}{y}\right)+{\cot }^{-1}\left(\frac{1}{z}\right)=3\pi$, then prove that $x+y+z=xyz$

Answer 1

${\sec }^{-1}\left(\sqrt{1+x^2}\right)+{\csc }^{-1}\left(\frac{\sqrt{1+y^2}}{y}\right)+{\cot }^{-1}\left(\frac{1}{z}\right)=3\pi$

$\Rightarrow {\cos }^{-1}\left(\frac{1}{\sqrt{1+}{x}^2}\right)+{\sin }^{-1}\left(\frac{y}{\sqrt{1+y^2}}\right)+{\tan }^{-1}\left(z\right)=3\pi$

$\Rightarrow {\tan }^{-1}x+{\tan }^{-1}y+{\tan }^{-1}z=3\pi$

$\Rightarrow {\tan }^{-1}\left(\frac{x+y+z-xyz}{1-xy-yz-zx}\right)=3\pi$

$\Rightarrow \tan 3\pi =\frac{x+y+z-xyz}{1-xy-yz-zx}=0$

$\Rightarrow x+y+z=xyz$