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# If $\tan^{-1}x+\tan^{-1}y=\frac{4\pi}{5},then\;\cot^{-1}x+\cot^{-1}y$ equals:

$(A)\quad\frac{\pi}{5}\quad(B)\quad\frac{2\pi}{5}\quad(C)\quad\frac{3}{5}\quad(D)\quad{\pi}$

Toolbox:
• $tan^{-1}x+cot^{-1}x=\large\frac{\pi}{2}$
Ans - (A) $\frac{\pi}{5}$
From the above formula we can write
$cot^{-1}x=\large\frac{\pi}{2}-tan^{-1}x\:and\:cot^{-1}y=\large\frac{\pi}{2}-tan^{-1}y$
$\Rightarrow\: cot^{-1}x+cot^{-1}y=\large\frac{\pi}{2}- tan^{-1}x+\large\frac{\pi}{2}- tan^{-1}y$
$=\pi-(tan^{-1}x+tan^{-1}y)=\pi-\large\frac{4\pi}{5}=\large\frac{\pi}{5}$

edited Mar 16, 2013