Answer
$ \dfrac{1-\cos{y}}{\sin{y}}$
Work Step by Step
$\tan{A} = \dfrac{\sin{A}}{\cos{A}}$
$\tan{A} = \dfrac{\sin{A}}{\cos{A}} \times \dfrac{\sin{A}}{\sin{A}}$
$\tan{A} = \dfrac{\sin^2{A}}{\cos{A} \sin{A}}$
$\because \sin^2{A} = \dfrac{1-\cos{2A}}{2} \hspace{30pt} \& \hspace{30pt} \cos{A}\sin{A} = \dfrac{1}{2}\sin{2A}$
$\therefore \tan{A} = \dfrac{1-\cos{2A}}{\sin{2A}}$
Replacing $A$ by $\dfrac{y}{2}$
$\tan{\dfrac{y}{2}} = \dfrac{1-\cos{y}}{\sin{y}}$
