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### Answers (1)

Answer:

$\frac{\pi}{5}$

Hint:

You must know the rules of inverse trigonometric function.

Given:

$\tan^{-1}x+\tan^{-1}y=\frac{4\pi}{5}$, then find $\cot^{-1}x+\cot^{-1}y$

Solution:

Using inverse rule,

$\tan^{-1}x+\cot^{-1}x=\frac{\pi}{2}$ and

$\tan^{-1}y+\cot^{-1}y=\frac{\pi}{2}$

Adding both equations,

\begin{aligned} &\Rightarrow \quad \tan ^{-1} x+\cot ^{-1} x+\tan ^{-1} y+\cot ^{-1} y=\frac{\pi}{2}+\frac{\pi}{2} \\\\ &\Rightarrow \quad \tan ^{-1} x+\tan ^{-1} y+\cot ^{-1} x+\cot ^{-1} y=\frac{\pi}{2}+\frac{\pi}{2} \\\\ &\Rightarrow \quad \frac{4 \pi}{5}+\cot ^{-1} x+\cot ^{-1} y=\pi \end{aligned}

\begin{aligned} \Rightarrow & \cot ^{-1} x+\cot ^{-1} y=\pi-\frac{4 \pi}{5} \\\\ \Rightarrow & \cot ^{-1} x+\cot ^{-1} y=\frac{5 \pi-4 \pi}{5} \\\\ \Rightarrow & \cot ^{-1} x+\cot ^{-1} y=\frac{\pi}{5} \end{aligned}

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