Double angle (or half angle, depending how you look at it) identities:
[tex]\cos^2\dfrac x2=\dfrac{1+\cos x}2[/tex]
[tex]\sin^2\dfrac x2=\dfrac{1-\cos x}2[/tex]
[tex]\implies\cot^2\dfrac x2=\dfrac{\cos^2\frac x2}{\sin^2\frac x2}=\dfrac{1+\cos x}{1-\cos x}[/tex]
So we have
[tex]\cot^215^\circ=\dfrac{1+\cos30^\circ}{1-\cos30^\circ}=\dfrac{1+\frac{\sqrt3}2}{1-\frac{\sqrt3}2}[/tex]
[tex]\implies\cot^215^\circ=7+4\sqrt3[/tex]
[tex]\implies\cot15^\circ=\sqrt{7+4\sqrt3}=2+\sqrt3[/tex]
Note that when taking the square root, we should take into account that that could yield two possible solutions, but we know [tex]\cos15^\circ>0[/tex] and [tex]\sin15^\circ>0[/tex], so it's also the case that [tex]\cot15^\circ>0[/tex].
Also, the reason we have equality in the last step can be explained like so:
[tex]7+4\sqrt3=4+4\sqrt3+3=4+4\sqrt3+(\sqrt3)^2=(2+\sqrt3)^2[/tex]
(not unlike the process used to complete the square)
