sqrt 3 ,{rm{cosec}},{20^o} - sec ,{20^o} = | vedclass

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Question

(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 2

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(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 }} $

$= \frac{{4\cos 50^\circ }}{{\sin 40^\circ }} = \frac{{4\sin 40^\circ }}{{\sin 40^\circ }} = 4$.

$\sqrt 3 \,{\rm{cosec}}\,{20^o} - \sec \,{20^o} = $

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