Question

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

• A. $x+y+z-xyz=0$
• B. $x+y+z+xyz=0$
• C. $xy+xz+zy+1=0$
• D. $xy+xz+zy-1=0$

Answer 1

The correct option is D

$xy+xz+zy-1=0$

Explanation for the correct option:

Use formula ${\tan }^{-1}A+{\tan }^{-1}B={\tan }^{-1}\left(\frac{A+B}{1-AB}\right)$

Given the equation: ${\tan }^{-1}x+{\tan }^{-1}y+{\tan }^{-1}z=\frac{\pi }{2}$,

Simplify as shown,

$\begin{array}{c}{\tan }^{-1}x+{\tan }^{-1}y+{\tan }^{-1}z=\frac{\pi }{2}\\ {\tan }^{-1}x+{\tan }^{-1}y=\frac{\pi }{2}-{\tan }^{-1}z\\ \Rightarrow {\tan }^{-1}\frac{x+y}{1-xy}={\cot }^{-1}z\\ \Rightarrow \frac{x+y}{1-xy}=\tan \left({\tan }^{-1}\frac{1}{z}\right)\\ \Rightarrow \frac{x+y}{1-xy}=\frac{1}{z}\\ \Rightarrow xz+zy=1-xy\\ \Rightarrow xy+xz+zy-1=0\end{array}$

Hence, the correct option is (D).