A calculator must not be used in this question.

1. Express $15^\circ$ as the difference of two special angles and apply the compound angle formula for tangent.

$\tan 15^\circ = \tan(60^\circ - 45^\circ)$
$\tan(60^\circ - 45^\circ) = \frac{\tan 60^\circ - \tan 45^\circ}{1 + \tan 60^\circ \tan 45^\circ}$

2. Substitute the exact values of the trigonometric ratios for the special angles.

$\tan 60^\circ = \sqrt{3}$
$\tan 45^\circ = 1$
$\tan 15^\circ = \frac{\sqrt{3} - 1}{1 + (\sqrt{3})(1)} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$

3. Rationalise the denominator by multiplying the numerator and denominator by the conjugate.

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

4. Simplify the expression to obtain the final result.

$= \frac{4 - 2\sqrt{3}}{2}$
$= 2 - \sqrt{3}$ (shown)

1. Find $\cot 15^\circ$ using the result from part (a)

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

2. Apply the Pythagorean identity for cosec

Using $\operatorname{cosec}^2 A = 1 + \cot^2 A$:
$\operatorname{cosec}^2 15^\circ = 1 + \cot^2 15^\circ = 1 + (2 + \sqrt{3})^2$

3. Expand and simplify to the required form

$(2 + \sqrt{3})^2 = 4 + 4\sqrt{3} + 3 = 7 + 4\sqrt{3}$
$\operatorname{cosec}^2 15^\circ = 1 + 7 + 4\sqrt{3} = 8 + 4\sqrt{3}$
So $a = 8$, $b = 4$.