Question
Find the values:
$\tan^{-1}\bigg[2\cos\bigg(2\sin^{-1}\frac{1}{2}\bigg)\bigg]$

Answer

$\tan^{-1}\bigg[2\cos\bigg(2\sin^{-1}\frac{1}{2}\bigg)\bigg]=\tan^{-1}\bigg[2\cos\bigg(2\sin^{-1}\sin\frac{\pi}{6}\bigg)\bigg]$
$=\tan^{-1}\bigg[2\cos\bigg(2\times\frac{\pi}{6}\bigg)\bigg]=\tan^{-1}\bigg[2\cos\frac{\pi}{3}\bigg]$
$=\tan^{-1}\bigg[2\times\frac{1}{2}\bigg]=\tan^{-1}1=\tan^{-1}\tan\frac{\pi}{4}=\frac{\pi}{4}$

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