Question:
The value of $\frac{\cos ^{3} 20^{\circ}-\cos ^{3} 70^{\circ}}{\sin ^{3} 70^{\circ}-\sin ^{3} 20^{\circ}}$ is
(a) $\frac{1}{2}$
(b) $\frac{1}{\sqrt{2}}$
(c) 1
(d) 2
Solution:
We have to evaluate the value. The formula to be used,
$a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)$
$a^{3}-b^{3}=(a-b)\left(a^{2}+b^{2}+a b\right)$
So,
$=\frac{\cos ^{3} 20^{\circ}-\cos ^{3} 70^{\circ}}{\sin ^{3} 70^{\circ}-\sin ^{3} 20^{\circ}}$
$=\frac{\left(\cos 20^{\circ}-\cos 70\right)\left(\cos ^{2} 20^{\circ}+\cos ^{2} 70+\cos 20^{\circ} \cos 70^{\circ}\right)}{\left(\sin 70^{\circ}-\sin 20^{\circ}\right)\left(\sin ^{2} 70^{\circ}+\sin ^{2} 20^{\circ}+\sin 70^{\circ} \sin 20^{\circ}\right)}$
Now using the properties of complementary angles,
So the answer is $(c)$ ]