Question

Evaluate ${\tan }^{-1}\mathrm{x}+{\tan }^{-1}\mathrm{y}+{\tan }^{-1}\frac{1-\mathrm{x}-\mathrm{y}-\mathrm{xy}}{1+\mathrm{x}+\mathrm{y}-\mathrm{xy}}$

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{4}$
• C. $\mathrm{\pi }$
• D. $\frac{\pi }{3}$

Answer 1

The correct option is B

$\frac{\pi }{4}$

Explanation for correct option:

Evaluating the given expression ${\tan }^{-1}\mathrm{x}+{\tan }^{-1}\mathrm{y}+{\tan }^{-1}\frac{1-\mathrm{x}-\mathrm{y}-\mathrm{xy}}{1+\mathrm{x}+\mathrm{y}-\mathrm{xy}}$

Formula to be used : We know that ${\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\left(\frac{x+y}{1-xy}\right)$ and ${\tan }^{-1}x-{\tan }^{-1}y={\tan }^{-1}\left(\frac{x-y}{1+xy}\right)$.

Simplifying the expression,

$$\begin{gathered} ={\tan }^{-1}\mathrm{x}+{\tan }^{-1}\mathrm{y}+{\tan }^{-1}\frac{1-\mathrm{x}-\mathrm{y}-\mathrm{xy}}{1+\mathrm{x}+\mathrm{y}-\mathrm{xy}} \\ ={\tan }^{-1}\left[\frac{\mathrm{x}+\mathrm{y}}{1-\mathrm{xy}}\right]+{\tan }^{-1}\left[\frac{1-\mathrm{xy}-(\mathrm{x}+\mathrm{y})}{1-\mathrm{xy}+(\mathrm{x}+\mathrm{y})}\right] \\ ={\tan }^{-1}\left[\frac{\mathrm{x}+\mathrm{y}}{1-\mathrm{xy}}\right]+{\tan }^{-1}\left[\frac{{\displaystyle \frac{(1-xy)-\left(x+y\right)}{1-xy}}}{{\displaystyle \frac{\left(1-xy\right)+\left(x+y\right)}{1-xy}}}\right] \\ ={\tan }^{-1}\left[\frac{\mathrm{x}+\mathrm{y}}{1-\mathrm{xy}}\right]+{\tan }^{-1}\left[\frac{{\displaystyle 1-\frac{x+y}{1-xy}}}{{\displaystyle 1+\frac{x+y}{1-xy}\times 1}}\right] \\ ={\tan }^{-1}\left[\frac{\mathrm{x}+\mathrm{y}}{1-\mathrm{xy}}\right]+{\tan }^{-1}\left(1\right)-{\tan }^{-1}\frac{x+y}{1-xy} \\ ={\tan }^{-1}\left(1\right) \\ =\frac{\mathrm{\pi }}{4} \end{gathered}$$

Hence, the correct answer is option (B).