Question

Verify the equality $\frac{1}{\cos {290}^{\circ }}+\frac{1}{\sqrt{3}\sin {250}^{\circ }}=\frac{4}{\sqrt{3}}$

Answer 1

$\frac{1}{\cos 290°}+\frac{1}{\sqrt{3}\sin 250°}$

$=\frac{\sqrt{3}\sin 250°+\cos 290°}{\cos 290°\sqrt{3}\sin 250°}$

$=2[\frac{\sqrt{3}(-\cos 20°)}{2}+\frac{1}{2}\cos (270°+20°\left)\right]\times \frac{1}{\sqrt{3}\sin 250°\cos 290°}$

$=2[\frac{\sqrt{3}(-\cos 20°)}{2}+\frac{1}{2}\sin 20°]\times \frac{1}{\sqrt{3}\sin 20°\cos 20°}$

$=-4\left[\cos 50°\right]\times \frac{1}{-\sqrt{3}\times \sin 40°}$

$=\frac{4}{\sqrt{3}}$