Question

If ${\tan }^{-1}x+{\tan }^{-1}y=\frac{\pi }{4}$, then

• A. $x+y+xy=1$
• B. $x+y-xy=1$
• C. $x+y+xy+1=0$
• D. $x+y-xy+1=0$

Answer 1

The correct option is A

$x+y+xy=1$

Explanation for the correct option:

Given the equation: ${\tan }^{-1}x+{\tan }^{-1}y=\frac{\pi }{4}........\left(i\right)$,

we know $\begin{array}{rcl}\tan \frac{\pi }{4}& =& 1\end{array}$

Apply the formula ${\tan }^{-1}A+{\tan }^{-1}B={\tan }^{-1}\left(\frac{A+B}{1-AB}\right)$ in the equation (i),

$\begin{array}{c}{\tan }^{-1}x+{\tan }^{-1}y=\frac{\pi }{4}\\ {\tan }^{-1}\frac{x+y}{1-xy}=\frac{\pi }{4}\\ \frac{x+y}{1-xy}=\tan \frac{\pi }{4}\\ \frac{x+y}{1-xy}=1\\ x+y=1-xy\\ x+y+xy=1\end{array}$

Hence, the correct option is (A).