MCQ
$\sqrt 3 \,{\rm{cosec}}\,{20^o} - \sec \,{20^o} = $
- A$2$
- B$\frac{{2\,\sin {{20}^o}}}{{\sin {{40}^o}}}$
- ✓$4$
- D$\frac{{4\,\sin {{20}^o}}}{{\sin {{40}^o}}}$
$ = \frac{{\sqrt 3 \cos 20^\circ - \sin 20^\circ }}{{\sin 20^\circ \cos 20^\circ }} $
$= \frac{{2\left[ {\frac{{\sqrt 3 }}{2}\cos 20^\circ - \frac{1}{2}\sin \,20^\circ } \right]}}{{\frac{2}{2}\sin 20^\circ \cos 20^\circ }}$
$ = \frac{{4\cos (20^\circ + 30^\circ )}}{{\sin 40^\circ }} $
$= \frac{{4\cos 50^\circ }}{{\sin 40^\circ }} = \frac{{4\sin 40^\circ }}{{\sin 40^\circ }} = 4$.
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