Calculate, without using tables $\;$ $\dfrac{\sin \alpha}{\sin^3 \alpha + 3 \cos^3 \alpha}$, if $\tan \alpha = 2$ and $\alpha$ is in the first quadrant.
Given: $\;$ $\tan \alpha = 2$ $\implies$ $\tan^2 \alpha = 4$
$\therefore \;$ $1 + \tan^2 \alpha = 1 + 4 = 5$
i.e. $\;$ $\sec^2 \alpha = 5$
i.e. $\;$ $\dfrac{1}{\cos^2 \alpha} = 5$ $\implies$ $\cos^2 \alpha = \dfrac{1}{5}$
$\implies$ $\cos \alpha = \dfrac{\pm 1}{\sqrt{5}}$
But since $\alpha$ is in the first quadrant, $\;$ $\therefore \;$ $\cos \alpha = \dfrac{+ 1}{\sqrt{5}}$
Then, $\;$ $\sin \alpha = \sqrt{1 - \cos^2 \alpha} = \sqrt{1 - \dfrac{1}{5}} = \dfrac{\pm 2}{\sqrt{5}}$
But since $\alpha$ is in the first quadrant, $\;$ $\therefore \;$ $\sin \alpha = \dfrac{+ 2}{\sqrt{5}}$
$\therefore \;$ The given expression $\;$ $\dfrac{\sin \alpha}{\sin^3 \alpha + 3 \cos^3 \alpha}$ $\;$ becomes
$\dfrac{\dfrac{2}{\sqrt{5}}}{\left(\dfrac{2}{\sqrt{5}}\right)^3 + 3 \left(\dfrac{1}{\sqrt{5}}\right)^3}$
$= \dfrac{\dfrac{2}{\sqrt{5}}}{\dfrac{8}{5 \sqrt{5}} + \dfrac{3}{5 \sqrt{5}}}$
$= \dfrac{2}{\sqrt{5}} \times \dfrac{5 \sqrt{5}}{11} = \dfrac{10}{11}$