$\sqrt 3 \,{\rm{cosec}}\,{20^o} - \sec \,{20^o} = $
$\sqrt 3 \,{\rm{cosec}}\,{20^o} - \sec \,{20^o} = $
- [IIT 1988]
- A
$2$
- B
$\frac{{2\,\sin {{20}^o}}}{{\sin {{40}^o}}}$
- C
$4$
- D
$\frac{{4\,\sin {{20}^o}}}{{\sin {{40}^o}}}$
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