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3.Trigonometrical Ratios, Functions and Identities
medium
$\sqrt 3 \,{\rm{cosec}}\,{20^o} - \sec \,{20^o} = $
A
$2$
B
$\frac{{2\,\sin {{20}^o}}}{{\sin {{40}^o}}}$
C
$4$
D
$\frac{{4\,\sin {{20}^o}}}{{\sin {{40}^o}}}$
(IIT-1988)
Solution
(c) $\sqrt 3 {\rm{cosec}}\,20^\circ – \sec 20^\circ = \frac{{\sqrt 3 }}{{\sin 20^\circ }} – \frac{1}{{\cos \,20^\circ }}$
$ = \frac{{\sqrt 3 \cos 20^\circ – \sin 20^\circ }}{{\sin 20^\circ \cos 20^\circ }} $
$= \frac{{2\left[ {\frac{{\sqrt 3 }}{2}\cos 20^\circ – \frac{1}{2}\sin \,20^\circ } \right]}}{{\frac{2}{2}\sin 20^\circ \cos 20^\circ }}$
$ = \frac{{4\cos (20^\circ + 30^\circ )}}{{\sin 40^\circ }} $
$= \frac{{4\cos 50^\circ }}{{\sin 40^\circ }} = \frac{{4\sin 40^\circ }}{{\sin 40^\circ }} = 4$.
Standard 11
Mathematics
