# Find the values of each of the following:    11. $\tan^{-1}\left[2\cos\left(2\sin^{-1}\frac{1}{2} \right ) \right ]$

Given equation:

$\tan^{-1}\left[2\cos\left(2\sin^{-1}\frac{1}{2} \right ) \right ]$

So, solving the inner bracket first, we take the value of $\sin x^{-1} \frac{1}{2} = x.$

Then we have,

$\sin x = \frac{1}{2} = \sin \left ( \frac{\pi}{6} \right )$

Therefore, we can write $\sin^{-1} \frac{1}{2} = \frac{\pi}{6}$.

$\tan^{-1}\left[2\cos\left(2\sin^{-1}\frac{1}{2} \right ) \right ] = \tan^{-1}\left[2\cos\left(2\times\frac{\pi}{6} \right ) \right ]$

$= \tan^{-1}\left[2\cos\left(\frac{\pi}{3} \right ) \right ] = \tan^{-1}\left[2\times\left(\frac{1}{2} \right ) \right ] = \tan^{-1}1 = \frac{\pi}{4}$.

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