frac{1}{cos 290^{circ}}+frac{1}{sqrt{3} sin 2 | vedclass

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(B) આપેલ પદાવલિ: $E = \frac{1}{\cos 290^{\circ}} + \frac{1}{\sqrt{3} \sin 250^{\circ}}$ સંલગ્ન ખૂણાઓનો ઉપયોગ કરતા: $\cos 290^{\circ} = \cos(270^{\circ

Vedclass · Accepted Answer

$\frac{4}{\sqrt{3}}$

Vedclass · Answer

(B) આપેલ પદાવલિ: $E = \frac{1}{\cos 290^{\circ}} + \frac{1}{\sqrt{3} \sin 250^{\circ}}$ સંલગ્ન ખૂણાઓનો ઉપયોગ કરતા: $\cos 290^{\circ} = \cos(270^{\circ} + 20^{\circ}) = \sin 20^{\circ}$ અને $\sin 250^{\circ} = \sin(270^{\circ} - 20^{\circ}) = -\cos 20^{\circ}$. તેથી,$E = \frac{1}{\sin 20^{\circ}} - \frac{1}{\sqrt{3} \cos 20^{\circ}}$ $E = \frac{\sqrt{3} \cos 20^{\circ} - \sin 20^{\circ}}{\sqrt{3} \sin 20^{\circ} \cos 20^{\circ}}$ અંશ અને છેદને $2$ વડે ગુણતા: $E = \frac{2(\frac{\sqrt{3}}{2} \cos 20^{\circ} - \frac{1}{2} \sin 20^{\circ})}{\frac{\sqrt{3}}{2} (2 \sin 20^{\circ} \cos 20^{\circ})}$ $E = \frac{2(\sin 60^{\circ} \cos 20^{\circ} - \cos 60^{\circ} \sin 20^{\circ})}{\frac{\sqrt{3}}{2} \sin 40^{\circ}}$ $E = \frac{2 \sin(60^{\circ} - 20^{\circ})}{\frac{\sqrt{3}}{2} \sin 40^{\circ}} = \frac{2 \sin 40^{\circ}}{\frac{\sqrt{3}}{2} \sin 40^{\circ}} = \frac{4}{\sqrt{3}}$

$\frac{1}{\cos 290^{\circ}}+\frac{1}{\sqrt{3} \sin 250^{\circ}} = $

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