sqrt 3 , cosec, 20^o - sec, 20^o&n | vedclass

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Question

$\frac{{\sqrt 3 }}{{\sin {{20}^0}}}\,\, - \,\frac{1}{{\cos {{20}^0}}}$ $=$ $\frac{{2\left( {\frac{{\sqrt 3 }}{2}} \right)\,\,\cos {{20}^0}\,\, -

Vedclass · Accepted Answer

$4$

Vedclass · Answer

$\frac{{\sqrt 3 }}{{\sin {{20}^0}}}\,\, - \,\frac{1}{{\cos {{20}^0}}}$ $=$ $\frac{{2\left( {\frac{{\sqrt 3 }}{2}} \right)\,\,\cos {{20}^0}\,\, - \,\,\sin {{20}^0}}}{{\frac{{\sin {{40}^0}}}{2}}}$ $=$ $\frac{{\sin {{80}^0}\, + \,\sin {{40}^0}\,\, - \,\,\sin {{20}^0}}}{{\frac{{\sin {{40}^0}}}{2}}}$ $\frac{{2\,\sin {{30}^0}\,\cos {{50}^0}\,\, + \,\,\sin {{40}^0}}}{{\frac{{\sin {{40}^0}}}{2}}}$ $= 4$

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

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