If $\tan^{-1} x + \tan^{-1} y = \frac{4\pi}{5}$, then $\cot^{-1} x + \cot^{-1} y$ is equal to

$\frac{\pi}{5}$

The correct answer is Option (1) → $\frac{\pi}{5}$ ##

We have, $\tan^{-1} x + \tan^{-1} y = \frac{4\pi}{5} \dots(i)$

Let $\tan^{-1} x = \alpha ⇒\tan \alpha = x$

Again, let $\tan^{-1} y = \beta ⇒ \tan \beta = y$

From Eq. (i), we get

$\alpha + \beta = \frac{4\pi}{5} \dots(ii)$

Now,

$\cot^{-1} x + \cot^{-1} y = \cot^{-1}(\tan \alpha) + \cot^{-1}(\tan \beta)$$

$= \cot^{-1} \left[ \cot \left( \frac{\pi}{2} - \alpha \right) \right] + \cot^{-1} \left[ \cot \left( \frac{\pi}{2} - \beta \right) \right]$

$= \frac{\pi}{2} - \alpha + \frac{\pi}{2} - \beta$

$= \pi - (\alpha + \beta)$

$= \pi - \frac{4\pi}{5} \quad [\text{from Eq. (ii)}]$

$= \frac{\pi}{5}$