Question

Write the function in the simplest form: $\tan^{-1}\bigg(\sqrt{\frac{1-\cos x}{1+\cos x}}\bigg),x<\pi$

Answer 1

$\tan^{-1}\bigg(\sqrt{\frac{1-\cos x}{1+\cos x}}\bigg),x<\pi$

= $tan^{-1}(\sqrt \frac{2\sin^2\frac{x}{2}}{2\cos^2\frac{x}{2}}$

=$\tan^{-1}\bigg(\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}\bigg)=\tan^{-1}\bigg(tan\frac{x}{2}\bigg)$
=$\frac{x}{2}$