frac{1}{sin 250^{circ}}+frac{sqrt{3}}{cos 290 | vedclass

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Question

(B) આપેલ પદાવલિ: $\frac{1}{\sin 250^{\circ}}+\frac{\sqrt{3}}{\cos 290^{\circ}}$$= \frac{1}{\sin(270^{\circ}-20^{\circ})} + \frac{\sqrt{3}}{\cos(270^{\

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) આપેલ પદાવલિ: $\frac{1}{\sin 250^{\circ}}+\frac{\sqrt{3}}{\cos 290^{\circ}}$$= \frac{1}{\sin(270^{\circ}-20^{\circ})} + \frac{\sqrt{3}}{\cos(270^{\circ}+20^{\circ})}$$= -\frac{1}{\cos 20^{\circ}} + \frac{\sqrt{3}}{\sin 20^{\circ}}$$= \frac{-\sin 20^{\circ} + \sqrt{3} \cos 20^{\circ}}{\cos 20^{\circ} \sin 20^{\circ}}$$= \frac{2 \left( -\frac{1}{2} \sin 20^{\circ} + \frac{\sqrt{3}}{2} \cos 20^{\circ} \right)}{\frac{1}{2} \sin 40^{\circ}}$$= \frac{2 (\sin 60^{\circ} \cos 20^{\circ} - \cos 60^{\circ} \sin 20^{\circ})}{\frac{1}{2} \sin 40^{\circ}}$$= \frac{2 \sin(60^{\circ}-20^{\circ})}{\frac{1}{2} \sin 40^{\circ}}$$= \frac{4 \sin 40^{\circ}}{\sin 40^{\circ}} = 4$તેથી,વિકલ્પ $B$ સાચો છે.

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

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