Question

If ${\sec }^{-1}\sqrt{1+x^2}+cose{c}^{-1}\frac{\sqrt{1+y^2}}{y}+{\cot }^{-1}\frac{1}{z}=\pi$ then x + y + z is equal to

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

Answer 1

Answer: (A)

Given : ${\sec }^{-1}\sqrt{1+x^2}+cose{c}^{-1}\frac{\sqrt{1+y^2}}{y}+{\cot }^{-1}\frac{1}{z}=\pi$
$\Rightarrow {\tan }^{-1}x+{\tan }^{-1}y+{\tan }^{-1}z=\pi$
$\Rightarrow {\tan }^{-1}\left(\frac{x+y+z-xyz}{1-xy-yz-zx}\right)=\pi$
$\Rightarrow \frac{x+y+z-xyz}{1-xy-yz-zx}=\tan \pi =0$
$\Rightarrow x+y+z=xyz$