The value of the expression $$\sqrt 3 \,{\text{cosec}}\,{\text{2}}{{\text{0}}^ \circ } - \sec {20^ \circ }$$     is equal to

Solution :
The given expression is $$\sqrt 3 \,{\text{cosec}}\,{\text{2}}{{\text{0}}^ \circ } - \sec {20^ \circ }$$
$$\eqalign{ & = \frac{{\sqrt 3 }}{{\sin {{20}^ \circ }}} - \frac{1}{{\cos {{20}^ \circ }}} = \frac{{\sqrt 3 \cos {{20}^ \circ } - \sin {{20}^ \circ }}}{{\sin {{20}^ \circ }\cos {{20}^ \circ }}} \cr & = 4\left[ {\frac{{\frac{{\sqrt 3 }}{2}\cos {{20}^ \circ } - \frac{1}{2}\sin {{20}^ \circ }}}{{2\sin {{20}^ \circ }\cos {{20}^ \circ }}}} \right] \cr & = 4\left[ {\frac{{\sin {{60}^ \circ }\cos {{20}^ \circ } - \cos {{60}^ \circ }\sin {{20}^ \circ }}}{{\sin \left( {2 \times {{20}^ \circ }} \right)}}} \right] \cr & = \frac{{4\sin \left( {{{60}^ \circ } - {{20}^ \circ }} \right)}}{{\sin {{40}^ \circ }}} \cr & = \frac{{4\sin {{40}^ \circ }}}{{\sin {{40}^ \circ }}} \cr & = 4 \cr} $$