sqrt{3} operatorname{cosec} 20^{circ} - sec 2 | vedclass

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Question

(B) આપેલ પદાવલિ $\sqrt{3} \operatorname{cosec} 20^{\circ} - \sec 20^{\circ}$ છે.$= \frac{\sqrt{3}}{\sin 20^{\circ}} - \frac{1}{\cos 20^{\circ}} = \fra

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) આપેલ પદાવલિ $\sqrt{3} \operatorname{cosec} 20^{\circ} - \sec 20^{\circ}$ છે.$= \frac{\sqrt{3}}{\sin 20^{\circ}} - \frac{1}{\cos 20^{\circ}} = \frac{\sqrt{3} \cos 20^{\circ} - \sin 20^{\circ}}{\sin 20^{\circ} \cos 20^{\circ}}$અંશ અને છેદને $2$ વડે ગુણતા:$= \frac{2 [(\frac{\sqrt{3}}{2}) \cos 20^{\circ} - (\frac{1}{2}) \sin 20^{\circ}]}{\sin 20^{\circ} \cos 20^{\circ}}$નિત્યસમ $\sin 2	heta = 2 \sin 	heta \cos 	heta$ નો ઉપયોગ કરતા,$\sin 20^{\circ} \cos 20^{\circ} = \frac{1}{2} \sin 40^{\circ}$ મળે.વળી,$\frac{\sqrt{3}}{2} = \sin 60^{\circ}$ અને $\frac{1}{2} = \cos 60^{\circ}$ છે.$= \frac{2 [\sin 60^{\circ} \cos 20^{\circ} - \cos 60^{\circ} \sin 20^{\circ}]}{\frac{1}{2} \sin 40^{\circ}}$સૂત્ર $\sin(A - B) = \sin A \cos B - \cos A \sin B$ નો ઉપયોગ કરતા:$= \frac{2 \sin(60^{\circ} - 20^{\circ})}{\frac{1}{2} \sin 40^{\circ}} = \frac{2 \sin 40^{\circ}}{\frac{1}{2} \sin 40^{\circ}} = 4$.

$\sqrt{3} \operatorname{cosec} 20^{\circ} - \sec 20^{\circ}$ ની કિંમત શોધો.

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