Question

If $x$ and $y$ are positive and $xy>1$, then what is ${\tan }^{-1}x+{\tan }^{-1}y$ equal to ?

• A. ${\tan }^{-1}\left(\frac{x+y}{1-xy}\right)$
• B. $\pi +{\tan }^{-1}\left(\frac{x+y}{1-xy}\right)$
• C. $\pi -{\tan }^{-1}\left(\frac{x+y}{1-xy}\right)$
• D. ${\tan }^{-1}\left(\frac{x-y}{1+xy}\right)$

Answer 1

The correct option is B $\pi +{\tan }^{-1}\left(\frac{x+y}{1-xy}\right)$

If $x>0,y>0$ such that $xy>1$,

then $x+y$ is positive and $1-xy$x+y1−xy is negative

Therefore, $\frac{x+y}{1-xy}$ is negative
x+y1−xy

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