If 90^{circ}

Question

(A) Given $\sin \alpha = \frac{\sqrt{3}}{2}$ where $90^{\circ} < \alpha < 180^{\circ}$ (Second Quadrant).Since $\sin \alpha$ is positive and $\cos \al

Vedclass · Accepted Answer

$\frac{2}{3}$

Vedclass · Answer

(A) Given $\sin \alpha = \frac{\sqrt{3}}{2}$ where $90^{\circ} Since $\sin \alpha$ is positive and $\cos \alpha$ is negative in the second quadrant,$\cos \alpha = -\sqrt{1 - \sin^2 \alpha} = -\sqrt{1 - \frac{3}{4}} = -\frac{1}{2}$.Thus,$	an \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{\sqrt{3}/2}{-1/2} = -\sqrt{3}$.Given $\sin \beta = -\frac{\sqrt{3}}{2}$ where $180^{\circ} Since $\sin \beta$ is negative and $\cos \beta$ is negative in the third quadrant,$\cos \beta = -\sqrt{1 - \sin^2 \beta} = -\sqrt{1 - \frac{3}{4}} = -\frac{1}{2}$.Thus,$	an \beta = \frac{\sin \beta}{\cos \beta} = \frac{-\sqrt{3}/2}{-1/2} = \sqrt{3}$.Substituting these values into the expression:$\frac{4 \sin \alpha - 3 	an \beta}{	an \alpha + \sin \beta} = \frac{4(\frac{\sqrt{3}}{2}) - 3(\sqrt{3})}{-\sqrt{3} + (-\frac{\sqrt{3}}{2})} = \frac{2\sqrt{3} - 3\sqrt{3}}{-\frac{3\sqrt{3}}{2}} = \frac{-\sqrt{3}}{-\frac{3\sqrt{3}}{2}} = \frac{2}{3}$.

If $90^{\circ} < \alpha < 180^{\circ}$,$\sin \alpha = \frac{\sqrt{3}}{2}$ and $180^{\circ} < \beta < 270^{\circ}$,$\sin \beta = -\frac{\sqrt{3}}{2}$,then $\frac{4 \sin \alpha - 3 \tan \beta}{\tan \alpha + \sin \beta} = $

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