Question:
If $\tan ^{-1} x+\tan ^{-1} y=\frac{5 \pi}{6}$, then $\cot ^{-1} x+\cot ^{-1} y=$ _________________.
Solution:
We know
$\tan ^{-1} a+\cot ^{-1} a=\frac{\pi}{2}$, for all $a \in \mathrm{R}$
Now,
$\tan ^{-1} x+\tan ^{-1} y=\frac{5 \pi}{6}$ (Given)
$\Rightarrow \frac{\pi}{2}-\cot ^{-1} x+\frac{\pi}{2}-\cot ^{-1} y=\frac{5 \pi}{6}$ [Using (1)]
$\Rightarrow \cot ^{-1} x+\cot ^{-1} y=\pi-\frac{5 \pi}{6}$
$\Rightarrow \cot ^{-1} x+\cot ^{-1} y=\frac{\pi}{6}$
If $\tan ^{-1} x+\tan ^{-1} y=\frac{5 \pi}{6}$, then $\cot ^{-1} x+\cot ^{-1} y=\frac{\pi}{6}$