Discussion Forum

The value of 

$\cos ^{-1}\left(\cot\left(\frac{\pi}{2}\right)\right)+\cos ^{-1}\left(\sin\left(\frac{2\pi}{3}\right)\right)$  is

Answer:

Explanation:

 We have, 

  $\cos ^{-1}\left(\cot\left(\frac{\pi}{2}\right)\right)+\cos ^{-1}\left(\sin\left(\frac{2\pi}{3}\right)\right)$ 

=$\cos ^{-1}(0)+\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)$

    $\left[\because \cot \frac{\pi}{2}=0  and  \sin \frac{2 \pi}{3}=\frac{\sqrt{3}}{2}\right]$

 =$cos ^{-1}\left( \cos \frac{\pi}{2}\right)+cos ^{-1}\left( \cos \frac{\pi}{6}\right)$

                  $\left[\because \cos  \frac{\pi}{2}=0   and  \cos \frac{\pi}{6}=\frac{\sqrt{3}}{2}\right]$

 = $\frac{\pi}{2}+\frac{\pi}{6}$

$=\frac{3\pi+\pi}{6}=\frac{4\pi}{6}=\frac{2 \pi}{3}$