Question
$\frac{1}{{\sin 10^\circ }} - \frac{{\sqrt 3 }}{{\cos 10^\circ }} =$
$ = \frac{{2\left( {\frac{{\cos 10^\circ }}{2} - \frac{{\sqrt 3 }}{2}\sin 10^\circ } \right)}}{{(2\sin 10^\circ \cos 10^\circ ) \times \frac{1}{2}}}$
$ = \frac{{4\,\sin \,({{30}^o} - {{10}^o})}}{{\sin \,{{20}^o}}} $
$= \frac{{4\,\sin \,{{20}^o}}}{{\sin \,{{20}^o}}} = 4$.
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$(A)$ $\left(\frac{9}{2 \sqrt{2}}, \frac{1}{\sqrt{2}}\right)$
$(B)$ $\left(-\frac{9}{2 \sqrt{2}},-\frac{1}{\sqrt{2}}\right)$
$(C)$ $(3 \sqrt{3},-2 \sqrt{2})$
$(D)$ $(-3 \sqrt{3}, 2 \sqrt{2})$