Without knowing what the possible answer choices are, perhaps this is one of them:
[tex]\tan^2\dfrac a2=\dfrac{\sin^2\frac a2}{\cos^2\frac a2}=\dfrac{\frac{1-\cos a}2}{\frac{1+\cos a}2}=\dfrac{1-\cos a}{1+\cos a}[/tex]
We can rewrite this in several ways, but one that should immediately occur to you is to consider writing the denominator in terms of [tex]\sin a[/tex]:
[tex]\dfrac{1-\cos a}{1+\cos a}\cdot\dfrac{1-\cos a}{1-\cos a}=\dfrac{1-2\cos a+\cos^2a}{1-\cos^2a}=\dfrac{1-2\cos a+\cos^2a}{\sin^2a}[/tex]
We can further write this in terms of the reciprocal functions,
[tex]\dfrac1{\sin^2a}-\dfrac{2\cos a}{\sin^2a}+\dfrac{\cos^2a}{\sin^2a}=\csc^2a-2\csc a\cot a+\cot^2a[/tex]
and so on...
