if

We know

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

$x<0, y<0$ such that

$x y=1$

Let $x=-a$ and $y=-b$ where both $a$ and $b$ are positive.

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

$=\tan ^{-1}\left(\frac{-a-a}{1-1}\right)$

$=\tan ^{-1}(-\infty)$

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

$=-\frac{\pi}{2}$