Question

If $x>0,y>0,z>0,xy+yz+zx<1$ and ${\tan }^{-1}x+{\tan }^{-1}y+{\tan }^{-1}z=\mathrm{\pi }$, then $x+y+z$ is equal to.

• A. $0$
• B. $xyz$
• C. $3xyz$
• D. $\sqrt{xyz}$

Answer 1

The correct option is B

$xyz$

Explanation for the correct answer:

To find the value of $x+y+z$:

Given,

$\begin{array}{rcl}{\tan }^{-1}x+{\tan }^{-1}y+{\tan }^{-1}z& =& \mathrm{\pi }\\ & \Rightarrow & {\tan }^{-1}\mathrm{x}+{\tan }^{-1}\mathrm{y}=\mathrm{\pi }-{\tan }^{-1}\mathrm{z}\\ & \Rightarrow & {\tan }^{-1}\frac{\left(\mathrm{x}+\mathrm{y}\right)}{1-\mathrm{xy}}={\tan }^{-1}\left(-\mathrm{z}\right)\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{[}\mathbf{\because }{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\mathbf{a}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\mathbf{b}\mathbf{=}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\frac{\left(a+b\right)}{\mathbf{1}\mathbf{-}\mathbf{ab}}\mathbf{,}\mathbf{}\mathbf{and}\mathbf{}\mathbf{\pi }\mathbf{-}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\mathbf{a}\mathbf{=}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\mathbf{-}\mathbf{a}\mathbf{]}\\ & \Rightarrow & \left(\mathrm{x}+\mathrm{y}\right)=\left(-\mathrm{z}\right)\left(1-\mathrm{xy}\right)\\ & \Rightarrow & \mathrm{x}+\mathrm{y}+\mathrm{z}=\mathrm{xyz}\end{array}$

Hence, the correct option is B.