Q.Prove the following: $\dfrac{1-\cos\theta}{1+\cos\theta}=\tan^2\dfrac{\theta}{2}$

A multiple angle is an integer multiple of a given angle, such as $2\theta$ (double) or $3\theta$ (triple); a submultiple angle is the given angle expressed as a multiple of a smaller one, such as writing $\theta$ as twice $\dfrac{\theta}{2}$. Every double- and triple-angle formula is obtained by treating $2\theta=\theta+\theta$ and $3\theta=2\theta+\theta$ and applying the compound-angle formulas of the sum-and-difference concept: $\sin2\theta=2\sin\theta\cos\theta$, $\cos2\theta=\cos^2\theta-\sin^2\theta=2\cos^2\theta-1=1-2\sin^2\theta$, $\tan2\theta=\dfrac{2\tan\theta}{1-\tan^2\theta}$, $\sin3\theta=3\sin\theta-4\sin^3\theta$, $\cos3\theta=4\cos^3\theta-3\cos\theta$ and $\tan3\theta=\dfrac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}$. Writing the same relations with $\theta$ replaced by $\dfrac{\theta}{2}$ turns them into …