Question

If ${\cot }^{-1}x+{\cot }^{-1}y+{\cot }^{-1}z=\frac{\pi }{2}$, then $x+y+z$ is also equal to

• A. $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$
• B. $xyz$
• C. $xy+yz+zx$
• D. None of these

Answer 1

The correct option is B $xyz$

We know that ${\cot }^{-1}x+{\cot }^{-1}y={\cot }^{-1}\frac{xy-1}{y+x}$

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

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

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

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

$\Rightarrow \frac{\left(\frac{xy-1}{y+x}\right)z-1}{z+\frac{xy-1}{y+x}}=0$

$\frac{xyz-z-x-y}{zy+x+xy-1}=0$

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

$\Rightarrow x+y+z=xyz$