Question

Prove that $\frac{1}{\sin {10}^{\circ }}-\frac{\sqrt{3}}{\cos {10}^{\circ }}=4$

Answer 1

$L.H.S=\frac{1\cdot cos{10}^{\circ }-\sqrt{3}sin{10}^{\circ }}{sin{10}^{\circ }cos{10}^{\circ }}$
put 1 = r cos $\alpha ,\sqrt{3}$ = r sin $\alpha$
$\therefore r^2=1+3=4,\therefore r=2$
and tan$\alpha =\sqrt{3},\therefore \alpha ={60}^{\circ }$
$L.H.S=\frac{r(cos{60}^{\circ }cos{10}^{\circ }-sin{60}^{\circ }sin{10}^{\circ })}{sin{10}^{\circ }cos{10}^{\circ }}$
$=\frac{2cos({60}^{\circ }+{10}^{\circ })}{\frac{1}{2}sin{20}^{\circ }}=4\cdot \frac{cos{70}^{\circ }}{sin{20}^{\circ }}=4\cdot \frac{sin{20}^{\circ }}{sin{20}^{\circ }}=4$