Question

Express the following in the simplest form
${\tan }^{-1}\left(\frac{\cos x}{1+\sin x}\right),\frac{-\pi }{2}<x<\frac{\pi }{2}$

Answer 1

${\tan }^{-1}\left(\frac{\cos x}{1+\sin x}\right),\frac{-\pi }{2}<2<\frac{\pi }{2}$

$⟹{\tan }^{-1}\left(\frac{{\cos }^{(}x/2)-{\sin }^{(}x/2)}{1-2\sin \left(x/2\right)\cos \left(x/2\right)}\right)$

$⟹{\tan }^{-1}\left(\frac{\cos \left(x/2\right)+\sin (x/2}{\cos \left(x/2\right)-\sin \left(x/2\right)}\right)$

Dividing numerator and denominator by $\cos \left(x/2\right)$

we get

${\tan }^{-1}\left(\frac{1+{\tan }^{-1}\left(x/2\right)}{1-{\tan }^{-1}\left(x/2\right)}\right)$

$⟹{\tan }^{-1}\left(\tan \right(\pi /4)+(x/2\left)\right)=\frac{\pi }{4}+\frac{x}{2}$