Question

Evaluate $\sqrt{3}cosec{20}^0-\sec {20}^0=$

Answer 1

L.H.S. $=\frac{\sqrt{3}}{\sin {20}^0}-\frac{1}{\cos {20}^0}=\frac{\sqrt{3}\cos {20}^0-\sin {20}^0}{\sin {20}^0\cos {20}^0}$
$=\frac{2(\frac{\sqrt{3}}{2}\cos {20}^0-\frac{1}{2}\sin {20}^0)}{\sin {20}^0\cos {20}^0}$
$=\frac{2(\sin {60}^0\cos {20}^0-\cos {60}^0\sin {20}^0)}{\sin {20}^0\cos {20}^0}$
$=\frac{2\sin ({60}^0-{20}^0)}{\sin {20}^0\cos {20}^0}=\frac{2\sin {40}^0}{\sin {20}^0\cos {20}^0}$
$=\frac{4\sin {40}^0}{2\sin {20}^0\cos {20}^0}=\frac{4\sin {40}^0}{\sin {40}^0}=4$
Ans: 4