We have
\(\sqrt3\) cosec20º - sec20º
= \(\frac{\sqrt3}{\sin20^o}\)- \(\frac{1}{\cos20^o}\)
= \(\frac{\sqrt3\cos20^o-\sin20^o}{\sin20^o\cos20^o}\)
= \(4\Bigg(\frac{\frac{\sqrt3}{2}\cos20^o-\frac{1}{2}\sin20^o}{2\sin20^o\cos20^o}\Bigg)\)
= \(4\Big(\frac{\sin60^o\cos20^o-\cos60^o\sin20^o}{\sin40^o}\Big)\)
= \(4\Big(\frac{\sin(60^o-20^o)}{\sin40^o}\Big)\) = 4
