If $x<0, y<0$ such that $x y=1$, then $\tan ^{-1} x+\tan ^{-1} y$ equals

Solution:
(b) $-\frac{\pi}{2}$
We know that $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$
$x<0, y<0$ such that
$x y=1$
Let $x=-a$ and $y=-b$, where $a$ and $b$ both are positive.
$\therefore \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$
$=\tan ^{-1}\left(\frac{-a-a}{1-1}\right)$
$=\tan ^{-1}(-\infty)$
$=\tan ^{-1}\left\{\tan \left(-\frac{\pi}{2}\right)\right\}$
$=-\frac{\pi}{2}$