A) \[xyz\]
B) \[2xyz\]
C) \[xy{{z}^{2}}\]
D) \[{{x}^{2}}yz\]
Correct Answer: A
Solution :
Given, \[{{\sec }^{-1}}\sqrt{1+{{x}^{2}}}+\cos e{{c}^{-1}}\] \[\frac{\sqrt{1+{{y}^{2}}}}{y}+{{\cot }^{-1}}\frac{1}{z}=\pi \] \[\therefore \] \[{{\tan }^{-1}}x+{{\tan }^{-1}}y+{{\tan }^{-1}}z=\pi \] \[\Rightarrow \]\[{{\tan }^{-1}}\left( \frac{x+y+z-xyz}{1-xy-yz-zx} \right)=\pi \] \[\Rightarrow \]\[x+y+z=xyz\]You need to login to perform this action.
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