Question

If ${\sec }^{-1}\sqrt{1+x^2}+{\text{cosec}}^{-1}\left(\frac{\sqrt{1+y^2}}{y}\right)+{\cot }^{-1}\left(\frac{1}{z}\right)=\pi$ , then $x+y+z$ is equal to :

• A. $xyz$
• B. $2xyz$
• C. $xy{z}^2$
• D. $x^{2}yz$

Answer 1

The correct option is A $xyz$

Given , ${\sec }^{-1}\sqrt{1+x^2}+{\text{cosec}}^{-1}\left(\frac{\sqrt{1+y^2}}{y}\right)+{\cot }^{-1}\left(\frac{1}{z}\right)=\pi$
$\Rightarrow {\tan }^{-1}x+{\tan }^{-1}y+{\tan }^{-1}z=\pi$

$\Rightarrow {\tan }^{-1}\left(\frac{x+y+z-xyz}{1-xy-yz-zx}\right)=\pi ={\tan }^{-1}\left(0\right)$

$\Rightarrow x+y+z-xyz=0$
$\Rightarrow x+y+z=xyz$