If cos theta = -frac{sqrt{3}}{2} and sin alph | vedclass

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(B) Given $\cos 	heta = -\frac{\sqrt{3}}{2} < 0$. Since $	heta$ does not lie in the third quadrant,$	heta$ must lie in the second quadrant. In the

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$\frac{5}{22}$

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(B) Given $\cos 	heta = -\frac{\sqrt{3}}{2} In the second quadrant,$	an 	heta = -\frac{1}{\sqrt{3}}$ and $\cot 	heta = -\sqrt{3}$. Thus,$\cot^2 	heta = (-\sqrt{3})^2 = 3$. Given $\sin \alpha = -\frac{3}{5}$. Since $\sin \alpha Case $1$: If $\alpha$ is in the third quadrant,$	an \alpha = \frac{3}{4}$ and $\cos \alpha = -\frac{4}{5}$. The expression becomes $\frac{2(\frac{3}{4}) + \sqrt{3}(-\frac{1}{\sqrt{3}})}{3 - \frac{4}{5}} = \frac{\frac{3}{2} - 1}{\frac{15-4}{5}} = \frac{1/2}{11/5} = \frac{5}{22}$. Case $2$: If $\alpha$ is in the fourth quadrant,$	an \alpha = -\frac{3}{4}$ and $\cos \alpha = \frac{4}{5}$. The expression becomes $\frac{2(-\frac{3}{4}) + \sqrt{3}(-\frac{1}{\sqrt{3}})}{3 + \frac{4}{5}} = \frac{-\frac{3}{2} - 1}{\frac{15+4}{5}} = \frac{-5/2}{19/5} = -\frac{25}{38}$. Since $\frac{5}{22}$ is the only option provided,the correct answer is $\frac{5}{22}$.

If $\cos \theta = -\frac{\sqrt{3}}{2}$ and $\sin \alpha = -\frac{3}{5}$,where $\theta$ does not lie in the third quadrant,then $\frac{2 \tan \alpha + \sqrt{3} \tan \theta}{\cot^2 \theta + \cos \alpha}$ is equal to

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