Question

$\frac{{\cos }^2{30}^{\circ }+\cos {30}^{\circ }\sin {30}^{\circ }+{\sin }^2{30}^{\circ }}{{\cos }^3{30}^{\circ }-{\sin }^3{30}^{\circ }}=?$

Answer 1

$\frac{{\cos }^{2}30°+\cos 30°\sin 30°+{\sin }^{2}30°}{{\cos }^{3}30°-{\sin }^{3}30°}$

Using identity: $a^3-b^3=(a-b)(a^2+ab+b^2)$

$\frac{{\cos }^{2}30°+\cos 30°\sin 30°+{\sin }^{2}30°}{{\cos }^{3}30°-{\sin }^{3}30°}$

$=\frac{{\cos }^{2}30°+\cos 30°\sin 30°+{\sin }^{2}30°}{{\cos }^{3}30°-{\sin }^{3}30°}$

$=\frac{({\cos }^{2}30°+\cos 30°\sin 30°+{\sin }^{2}30°)}{(\cos 30°-\sin 30°)({\cos }^{2}30°+\cos 30°\sin 30°+{\sin }^{2}30°)}$

$=\frac{1}{\cos 30°-\sin 30°}$

From trigonometric table-

$\cos 30°=\frac{\sqrt{3}}{2}$

$\sin 30°=\frac{1}{2}$

$\frac{1}{\cos 30°-\sin 30°}$

$=\frac{1}{\frac{\sqrt{3}}{2}-\frac{1}{2}}$

$=\frac{2}{\sqrt{3}-1}$

$=\frac{2}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}$

$=\frac{2(\sqrt{3}+1)}{3-1}$

$=\sqrt{3}+1$

Hence

$\frac{{\cos }^{2}30°+\cos 30°\sin 30°+{\sin }^{2}30°}{{\cos }^{3}30°-{\sin }^{3}30°}=\sqrt{3}+1$

Hence the correct answer is $\sqrt{3}+1$.