Question

If $A={30}^0$;

$\frac{{\cos }^{3}A-\cos 3A}{\cos A}+\frac{{\sin }^{3}A+\sin 3A}{\sin A}=3$

If true enter 1 else 0

Answer 1

Given, $A={30}^{\circ }$
To prove: $\frac{{\cos }^{3}A-\cos 3A}{\cos A}+\frac{{\sin }^{3}A+\sin 3A}{\sin A}=3$
L.H.S: $\frac{{\cos }^{3}A-\cos 3A}{\cos A}+\frac{{\sin }^{3}A+\sin 3A}{\sin A}$
= $\frac{{\cos }^3\left({30}^0\right)-\cos {90}^0}{\cos {30}^0}+\frac{{\sin }^3\left(30\right)+\sin 90}{\sin {30}^0}$
= $\frac{(\frac{\sqrt{3}}{2}{)}^3}{\frac{\sqrt{3}}{2}}+\frac{(\frac{1}{2}{)}^3+1}{\frac{1}{2}}$
= $\frac{3}{4}+\frac{\frac{9}{8}}{\frac{1}{2}}$
= $\frac{3}{4}+\frac{9}{4}$
= $\frac{12}{4}$
= $3$