Question

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

• A. $\pi$
• B. $\frac{\pi}{5}$
• C. $\frac{2 \pi}{5}$
• D. $\frac{3 \pi}{5}$

Answer 1

Answer: (B)

$\frac{\pi}{5}$

Answer 2

The correct answer is B:$\frac{\pi}{5}$
Given that;
$tan^{-1}x+tan^{-1}y=\frac{4\pi}{5}$-(i)
we know that,
$tan^{-1}x+cot^{-1}x=\frac{\pi}{2}$
$(\therefore tan^{-1}=\frac{\pi}{2}-cot^{-1}x)$
Then on converting the parent equation to the above form we get;
$\frac{\pi}{2}-cot^{-1}x+\frac{\pi}{2}-cot^{-1}y=\frac{4\pi}{5}$
⇒$-cot^{-1}x-cot^{-1}y=\frac{4\pi}{5}-\pi$
⇒$cot^{-1}x+cot^{-1}y=\frac{\pi}{5}$