sin 20^{circ}(4+sec 20^{circ})= | vedclass

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Question

(A) Given expression: $\sin 20^{\circ}(4+\sec 20^{\circ})$$= \sin 20^{\circ} \left(4 + \frac{1}{\cos 20^{\circ}}\right)$$= \frac{4 \sin 20^{\circ} \co

Vedclass · Accepted Answer

$\sqrt{3}$

Vedclass · Answer

(A) Given expression: $\sin 20^{\circ}(4+\sec 20^{\circ})$$= \sin 20^{\circ} \left(4 + \frac{1}{\cos 20^{\circ}}\right)$$= \frac{4 \sin 20^{\circ} \cos 20^{\circ} + \sin 20^{\circ}}{\cos 20^{\circ}}$$= \frac{2(2 \sin 20^{\circ} \cos 20^{\circ}) + \sin 20^{\circ}}{\cos 20^{\circ}}$$= \frac{2 \sin 40^{\circ} + \sin 20^{\circ}}{\cos 20^{\circ}}$$= \frac{2 \sin(60^{\circ} - 20^{\circ}) + \sin 20^{\circ}}{\cos 20^{\circ}}$$= \frac{2(\sin 60^{\circ} \cos 20^{\circ} - \cos 60^{\circ} \sin 20^{\circ}) + \sin 20^{\circ}}{\cos 20^{\circ}}$$= \frac{2(\frac{\sqrt{3}}{2} \cos 20^{\circ} - \frac{1}{2} \sin 20^{\circ}) + \sin 20^{\circ}}{\cos 20^{\circ}}$$= \frac{\sqrt{3} \cos 20^{\circ} - \sin 20^{\circ} + \sin 20^{\circ}}{\cos 20^{\circ}}$$= \frac{\sqrt{3} \cos 20^{\circ}}{\cos 20^{\circ}} = \sqrt{3}$

List-$A$	List-$B$
$(I)$ $\tan \left(\frac{\alpha + \beta}{2}\right) =$	$(a)$ $\frac{b}{a}$
$(II)$ $\cos (\alpha + \beta) =$	$(b)$ $\frac{2ab}{a^2 + b^2}$
$(III)$ $\sin (\alpha + \beta) =$	$(c)$ $\frac{2ab}{a^2 - b^2}$
$(IV)$ $\tan (\alpha + \beta) =$	$(d)$ $\frac{a^2 - b^2}{a^2 + b^2}$

$\sin 20^{\circ}(4+\sec 20^{\circ})=$

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