# Show prove the below identity? 1/cos290 + 1/(sqrt3sin250) = 4/sqrt3

## $\frac{1}{\cos} 290 + \frac{1}{\sqrt{3} \sin 250} \equiv \frac{4}{\sqrt{3}}$

May 11, 2018

$L H S = \frac{1}{\cos {290}^{\circ}} + \frac{1}{\sqrt{3} \sin {250}^{\circ}}$

$= \frac{1}{\cos {\left(360 - 70\right)}^{\circ}} + \frac{1}{\sqrt{3} \sin {\left(180 + 70\right)}^{\circ}}$

$= \frac{1}{\cos {70}^{\circ}} - \frac{1}{\sqrt{3} \sin {70}^{\circ}}$

$= \frac{\sqrt{3} \sin {70}^{\circ} - \cos {70}^{\circ}}{\sqrt{3} \sin {70}^{\circ} \cos {70}^{\circ}}$

$= \frac{1}{\sqrt{3}} \left[\frac{2 \left\{\sqrt{3} \sin {70}^{\circ} - \cos {70}^{\circ}\right\}}{2 \sin {70}^{\circ} \cos {70}^{\circ}}\right]$

$= \frac{1}{\sqrt{3}} \left[\frac{2 \cdot 2 \left\{\sin {70}^{\circ} \cdot \left(\frac{\sqrt{3}}{2}\right) - \cos {70}^{\circ} \cdot \left(\frac{1}{2}\right)\right\}}{\sin {140}^{\circ}}\right]$

$= \frac{1}{\sqrt{3}} \left[\frac{4 \left\{\sin {70}^{\circ} \cdot \cos {30}^{\circ} - \cos {70}^{\circ} \cdot \sin {30}^{\circ}\right\}}{\sin {\left(180 - 40\right)}^{\circ}}\right]$

$= \frac{1}{\sqrt{3}} \left[\frac{4 \left\{\sin {\left(70 - 30\right)}^{\circ}\right\}}{\sin {40}^{\circ}}\right] = \frac{1}{\sqrt{3}} \left[\frac{4 \left\{\cancel{\sin {40}^{\circ}}\right\}}{\cancel{\left(\sin {40}^{\circ}\right)}}\right] = \frac{4}{\sqrt{3}} = R H S$

NOTE that $\cos {\left(360 - A\right)}^{\circ} = \cos A \mathmr{and} \sin {\left(180 + A\right)}^{\circ} = - \sin A$

May 11, 2018

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

$= \frac{1}{\cos} \left(270 + 20\right) + \frac{1}{\sqrt{3} \sin \left(270 - 20\right)}$

$= \frac{1}{\sin} 20 - \frac{1}{\sqrt{3} \cos 20}$

$= \left[\frac{\sqrt{3} \cos 20 - \sin 20}{\sqrt{3} \sin 20 \cos 20}\right]$

$= \frac{2}{\sqrt{3}} \left[\frac{\frac{\sqrt{3}}{2} \cos 20 - \frac{1}{2} \sin 20}{\sin 20 \cos 20}\right]$

$= \frac{4}{\sqrt{3}} \left[\frac{\sin 60 \cos 20 - \cos 60 \sin 20}{2 \sin 20 \cos 20}\right]$

$= \frac{4}{\sqrt{3}} \left[\sin \frac{60 - 20}{2 \sin 20 \cos 20}\right]$

$= \frac{4}{\sqrt{3}} \left[\sin \frac{40}{\sin} 40\right]$

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