Answer
The required solution is $\underline{\tan \frac{\alpha }{2}=\frac{1-\cos \alpha }{\sin \alpha }}$
Work Step by Step
We know that the half-angle formula gives the relationship between angle $\alpha,$ and angle $\frac{\alpha }{2}.$ The described half-angle formula for tangent can be verified from the right-hand side. That means from $\frac{1-\cos \alpha }{\sin \alpha }$.
So, the expression for $\cos \alpha $ and $\sin \alpha $ can be obtained from the double-angle formula and replacing angle $\theta $ with angle $\frac{\alpha }{2}.$ This means, we start with the double-angle formula,
$\begin{align}
& \cos 2\theta =1-2si{{n}^{2}}\theta \\
& \sin 2\theta =2\sin \theta \cos \theta
\end{align}$
Then, replace angle $\theta $ with angle $\frac{\alpha }{2}.\text{ Then,}$
$\begin{align}
& \cos \alpha =1-2si{{n}^{2}}\frac{\alpha }{2} \\
& \sin \alpha =2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}
\end{align}$
Now, substitute these values in
$\frac{1-\cos \alpha }{\sin \alpha }$
$\begin{align}
& \frac{1-\cos \alpha }{\sin \alpha }=\frac{1-\left( 1-2{{\sin }^{2}}\frac{\alpha }{2} \right)}{2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}} \\
& =\frac{2{{\sin }^{2}}\frac{\alpha }{2}}{2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}} \\
& =\frac{\sin \frac{\alpha }{2}}{\cos \frac{\alpha }{2}} \\
& =\tan \frac{\alpha }{2}
\end{align}$
Thus, the half-angle formula for tangent is verified.
