frac{sin ^2(-160^{circ})}{sin ^2 70^{circ}} + | vedclass

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Question

(A) Given expression: $\frac{\sin ^2(-160^{\circ})}{\sin ^2 70^{\circ}} + \frac{\sin (180^{\circ}-	heta)}{\sin 	heta}$ Since $\sin(-\alpha) = -\sin(

Vedclass · Accepted Answer

$\sec ^2(20^{\circ})$

Vedclass · Answer

(A) Given expression: $\frac{\sin ^2(-160^{\circ})}{\sin ^2 70^{\circ}} + \frac{\sin (180^{\circ}-	heta)}{\sin 	heta}$ Since $\sin(-\alpha) = -\sin(\alpha)$,we have $\sin^2(-160^{\circ}) = (-\sin 160^{\circ})^2 = \sin^2 160^{\circ}$. Also,$\sin(180^{\circ}-	heta) = \sin 	heta$. So,the expression becomes $\frac{\sin^2 160^{\circ}}{\sin^2 70^{\circ}} + \frac{\sin 	heta}{\sin 	heta} = \frac{\sin^2(180^{\circ}-20^{\circ})}{\sin^2 70^{\circ}} + 1$. Since $\sin(180^{\circ}-\alpha) = \sin \alpha$,$\sin 160^{\circ} = \sin 20^{\circ}$. Thus,$\frac{\sin^2 20^{\circ}}{\sin^2 70^{\circ}} + 1 = \frac{\sin^2 20^{\circ}}{\cos^2 20^{\circ}} + 1 = 	an^2 20^{\circ} + 1 = \sec^2 20^{\circ}$.

$\frac{\sin ^2(-160^{\circ})}{\sin ^2 70^{\circ}} + \frac{\sin (180^{\circ}-\theta)}{\sin \theta} = $

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