# The result

Question:

The result $\tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)$ is true when value of $x y$ is __________________.

Solution:

We know

$\tan ^{-1} x-\tan ^{-1} y= \begin{cases}\tan ^{-1}\left(\frac{x-y}{1+x y}\right), & \text { if } x y>-1 \\ \pi+\tan ^{-1}\left(\frac{x-y}{1+x y}\right), & \text { if } x>0, y<0, x y<-1 \\ -\pi+\tan ^{-1}\left(\frac{x-y}{1+x y}\right), & \text { if } x<0, y>0, x y<-1\end{cases}$

Thus, $\tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)$ when the value of $x y>-1$.

The result $\tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)$ is true when value of $x y$ is __greater than − 1___.