Question

If ${\cot }^{-1}x+{\cot }^{-1}y+{\cot }^{-1}z=\frac{\pi }{2}$ then $x+y+z$ equals

• A. $xyz$
• B. $xy+yz+zx$
• C. $2xyz$
• D. None of these

Answer 1

The correct option is A $xyz$

${\cot }^{-1}x+{\cot }^{-1}y=\frac{\pi }{2}-{\cot }^{-1}z$
$\tan ({\cot }^{-1}x+{\cot }^{-1}y)=\tan (\frac{\pi }{2}-{\cot }^{-1}z)$
$\frac{\tan \left({\cot }^{-1}x\right)+\tan \left({\cot }^{-1}y\right)}{1-\tan \left({\cot }^{-1}x\right)\tan \left({\cot }^{-1}y\right)}=z$
$\frac{\frac{1}{x}+\frac{1}{y}}{1-\frac{1}{xy}}=z$
$\frac{x+y}{xy-1}=z$
$x+y+z=xyz$