Without using a calculator, show that $\cot 15^\circ = 2 + \sqrt{3}$. Hence show that $\csc^2 15^\circ = 4 \cot 15^\circ$.

1. Express $\cot 15^\circ$ in terms of $\tan 15^\circ$ and rewrite $15^\circ$ using special angles.

$\cot 15^\circ = \frac{1}{\tan 15^\circ}$
$\tan 15^\circ = \tan(45^\circ - 30^\circ)$

2. Apply the compound angle formula for tangent.

$\tan(45^\circ - 30^\circ) = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ}$

3. Substitute exact values for special angles.

$\tan 45^\circ = 1$, $\tan 30^\circ = \frac{1}{\sqrt{3}}$
$\tan 15^\circ = \frac{1 - \frac{1}{\sqrt{3}}}{1 + (1)(\frac{1}{\sqrt{3}})} = \frac{\frac{\sqrt{3} - 1}{\sqrt{3}}}{\frac{\sqrt{3} + 1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$

4. Find $\cot 15^\circ$ by taking the reciprocal and rationalising the denominator.

$\cot 15^\circ = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}$
$\cot 15^\circ = \frac{(\sqrt{3} + 1)(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)}$
$= \frac{3 + 2\sqrt{3} + 1}{3 - 1}$
$= \frac{4 + 2\sqrt{3}}{2}$
$= 2 + \sqrt{3}$

1. Express $\csc^2 15^\circ$ in terms of $\cot 15^\circ$ using the Pythagorean identity.

$\csc^2 15^\circ = 1 + \cot^2 15^\circ$

2. Substitute the value of $\cot 15^\circ$ from part (i) and simplify the left-hand side.

$\csc^2 15^\circ = 1 + (2 + \sqrt{3})^2$
$= 1 + (4 + 4\sqrt{3} + 3)$
$= 8 + 4\sqrt{3}$

3. Evaluate the right-hand side and conclude the proof.

$4 \cot 15^\circ = 4(2 + \sqrt{3})$
$= 8 + 4\sqrt{3}$

Since LHS = RHS, $\csc^2 15^\circ = 4 \cot 15^\circ$ (shown).