Evaluate :

$\dfrac{\text{cos 3 A - 2 cos 4 A}}{\text{sin 3 A + 2 sin 4 A}}$, when A = 15°.

$\dfrac{\text{cos 3 A - 2 cos 4 A}}{\text{sin 3 A + 2 sin 4 A}}\\[1em] = \dfrac{\text{cos (3 x 15°) - 2 cos (4 x 15°)}}{\text{sin (3 x 15°) + 2 sin (4 x 15°)}}\\[1em] = \dfrac{\text{cos 45° - 2 cos 60°}}{\text{sin 45° + 2 sin 60°}}\\[1em] = \dfrac{\Big(\dfrac{1}{\sqrt2}\Big) - 2 \times \Big(\dfrac{1}{2}\Big)}{\Big(\dfrac{1}{\sqrt2}\Big) + 2 \times \Big(\dfrac{\sqrt3}{2}\Big)}\\[1em] = \dfrac{\dfrac{1}{\sqrt2} - 1}{\dfrac{1}{\sqrt2} + \sqrt3}\\[1em] = \dfrac{\dfrac{1}{\sqrt2} - \dfrac{1 \times \sqrt2}{\sqrt2}}{\dfrac{1}{\sqrt2} + \dfrac{\sqrt3 \times \sqrt2}{\sqrt2}}\\[1em] = \dfrac{\dfrac{1}{\sqrt2} - \dfrac{\sqrt2}{\sqrt2}}{\dfrac{1}{\sqrt2} + \dfrac{\sqrt6}{\sqrt2}}\\[1em] = \dfrac{\dfrac{1 - \sqrt2}{\sqrt2}}{\dfrac{1 + \sqrt6}{\sqrt2}}\\[1em] = \dfrac{\dfrac{1 - \sqrt2}{\cancel{\sqrt2}}}{\dfrac{1 + \sqrt6}{\cancel{\sqrt2}}}\\[1em] = \dfrac{1 - \sqrt2}{1 + \sqrt6}\\[1em] = \dfrac{(1 - \sqrt2) \times (1 - \sqrt6)}{(1 + \sqrt6) \times (1 - \sqrt6)}\\[1em] = \dfrac{1 - \sqrt2 - \sqrt6 + \sqrt{12}}{1 - 6}\\[1em] = \dfrac{1 - \sqrt2 - \sqrt6 + 2\sqrt3}{-5}\\[1em] = \dfrac{1}{5}(- 1 + \sqrt2 + \sqrt6 - 2\sqrt3)$

Hence, $\dfrac{\text{cos 3 A - 2 cos 4 A}}{\text{sin 3 A + 2 sin 4 A}} = \dfrac{1}{5}(- 1 + \sqrt2 + \sqrt6 - 2\sqrt3)$