Question

$x+y=1,xy=2$, then ${\tan }^{-1}x+{\tan }^{-1}y=$

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

Answer 1

The correct option is B

$\frac{3\mathrm{\pi }}{4}$

Explanation for the correct option:

Calculate the value of the expression ${\tan }^{-1}x+{\tan }^{-1}y$

It is given that,

$x+y=1$ and $xy=2$

We know that, if $x>0,y>0$ and $xy>1$ then

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

$\Rightarrow$ ${\tan }^{-1}x+{\tan }^{-1}y=\pi +{\tan }^{-1}\left(\frac{1}{1-2}\right)$

$\Rightarrow$ ${\tan }^{-1}x+{\tan }^{-1}y=\pi +{\tan }^{-1}\left(-1\right)$

$\Rightarrow$ ${\tan }^{-1}x+{\tan }^{-1}y=\pi +\left(-\frac{\pi }{4}\right)$ $\left[\because {\tan }^{-1}\left(-1\right)=-\frac{\pi }{4}\right]$

$\Rightarrow$ ${\tan }^{-1}x+{\tan }^{-1}y=\frac{4\pi -\pi }{4}$

$\Rightarrow$ ${\tan }^{-1}x+{\tan }^{-1}y=\frac{3\pi }{4}$

Hence, the correct option is $B)\frac{3\pi }{4}$.