If $$tan \alpha=2$$, then the value of $$\frac{sin\alpha}{sin^3\alpha+cos^3\alpha}$$ is

Given : $$tan \alpha=2$$ ----------(i)

=> $$\frac{sin\alpha}{cos\alpha}=2$$

=> $$\frac{\sqrt{1-cos^2\alpha}}{cos\alpha}=2$$

=> $$\sqrt{1-cos^2\alpha}=2cos\alpha$$

Squaring both sides,

=> $$1-cos^2\alpha=4cos^2\alpha$$

=> $$4cos^2\alpha+1cos^2\alpha=1$$

=> $$cos^2\alpha=\frac{1}{5}$$ ----------(ii)

=> $$sin^2\alpha=1-\frac{1}{5}=\frac{4}{5}$$ ------------(iii)

To find : $$\frac{sin\alpha}{sin^3\alpha+cos^3\alpha}$$

Dividing both numerator and denominator by $$cos\alpha$$

= $$\frac{\frac{sin\alpha}{cos\alpha}}{\frac{sin^3\alpha}{cos\alpha}+\frac{cos^3\alpha}{cos\alpha}}$$

= $$\frac{tan\alpha}{tan\alpha .sin^2\alpha+cos^2\alpha}$$

Substituting values from equations (i),(ii) and (iii),

= $$\frac{2}{2\times\frac{4}{5}+\frac{1}{5}}$$

= $$\frac{2}{\frac{9}{5}}=\frac{10}{9}$$

=> Ans - (C)