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

Solving,

$\Rightarrow \dfrac{\cos 3A + 2 \cos 4A}{\sin 3A + 2\sin 4A}\\[1em] = \dfrac{\cos 3(15^\circ) + 2 \cos 4(15^\circ)}{\sin 3(15^\circ) + 2\sin 4(15^\circ)}\\[1em] = \dfrac{\cos 45^\circ + 2 \cos 60^\circ}{\sin 45^\circ + 2\sin 60^\circ}\\[1em] = \dfrac{\dfrac{1}{\sqrt{2}}+ 2\times\dfrac{1}{2}}{\dfrac{1}{\sqrt{2}}+ 2\times\dfrac{\sqrt{3}}{2}}\\[1em] = \dfrac{\dfrac{1}{\sqrt{2}}+ 1 }{\dfrac{1}{\sqrt{2}}+ {\sqrt{3}}}\\[1em] = \dfrac{\dfrac{1 + \sqrt{2}}{\sqrt{2}}}{\dfrac{1 + \sqrt{6}}{\sqrt{2}}}\\[1em] = \dfrac{1 + \sqrt{2}}{1 + \sqrt{6}}.$

Hence, $\dfrac{\cos 3A + 2 \cos 4A}{\sin 3A + 2\sin 4A} = \dfrac{1 + \sqrt{2}}{1 + \sqrt{6}}$.