L.H.S S . & =3 cosec 20^ - 20^ & = 3 20^ -1 20^ & =2 ( 3 2 20^ - 1 2 20^ ) & =2 ( 60^ 20^ - 60^ 20^ ) & =2 ( 60^ 20^ - 60^ 20^ 20^ 20^ ) & = 2 (60^ -20^ ) 1 2 (2 20^ 20^ ) & = 4 40^ 40^ & =4 & = R.H.S.

3 cosec 20^ - 20^ =4

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Question

$\sqrt{ } 3 \operatorname{cosec} 20^{\circ}-\sec 20^{\circ}=4$

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Answer

L.H.S
$\begin{aligned}
\mathrm{S} . & =\sqrt{3} \operatorname{cosec} 20^{\circ}-\sec 20^{\circ} \\
& =\frac{\sqrt{3}}{\sin 20^{\circ}}-\frac{1}{\cos 20^{\circ}} \\
& =2\left(\frac{\frac{\sqrt{3}}{2}}{\sin 20^{\circ}}-\frac{\frac{1}{2}}{\cos 20^{\circ}}\right)
\end{aligned}$
$\begin{aligned}
& =2\left(\frac{\sin 60^{\circ}}{\sin 20^{\circ}}-\frac{\cos 60^{\circ}}{\cos 20^{\circ}}\right) \\
& =2\left(\frac{\sin 60^{\circ} \cos 20^{\circ}-\cos 60^{\circ} \sin 20^{\circ}}{\sin 20^{\circ} \cos 20^{\circ}}\right) \\
& =\frac{2 \sin \left(60^{\circ}-20^{\circ}\right)}{\frac{1}{2}\left(2 \sin 20^{\circ} \cos 20^{\circ}\right)} \\
& =\frac{4 \sin 40^{\circ}}{\sin 40^{\circ}} \\
& =4\\
& =\text { R.H.S. }
\end{aligned}$

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