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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 $\cot ^{-1} x+\cot ^{-1} y=\frac{\pi}{6}$

 

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