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(C) Given that $A$ lies in the second quadrant,where $\cos A = -\frac{\sqrt{3}}{2}$.Since $\sin^2 A + \cos^2 A = 1$,we have $\sin A = \sqrt{1 - (-\fra

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$\frac{5}{22}$

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(C) Given that $A$ lies in the second quadrant,where $\cos A = -\frac{\sqrt{3}}{2}$.Since $\sin^2 A + \cos^2 A = 1$,we have $\sin A = \sqrt{1 - (-\frac{\sqrt{3}}{2})^2} = \sqrt{1 - \frac{3}{4}} = \frac{1}{2}$.Thus,$	an A = \frac{\sin A}{\cos A} = \frac{1/2}{-\sqrt{3}/2} = -\frac{1}{\sqrt{3}}$ and $\cot A = -\sqrt{3}$.Given that $B$ lies in the third quadrant,where $\sin B = -\frac{3}{5}$.Since $\sin^2 B + \cos^2 B = 1$,we have $\cos B = -\sqrt{1 - (-\frac{3}{5})^2} = -\sqrt{1 - \frac{9}{25}} = -\frac{4}{5}$.Thus,$	an B = \frac{\sin B}{\cos B} = \frac{-3/5}{-4/5} = \frac{3}{4}$.Now,substitute these values into the expression:$\frac{2 	an B + \sqrt{3} 	an A}{\cot^2 A + \cos B} = \frac{2(\frac{3}{4}) + \sqrt{3}(-\frac{1}{\sqrt{3}})}{(-\sqrt{3})^2 + (-\frac{4}{5})}$$= \frac{\frac{3}{2} - 1}{3 - \frac{4}{5}} = \frac{1/2}{11/5} = \frac{1}{2} 	imes \frac{5}{11} = \frac{5}{22}$.

If $A$ lies in the second quadrant and $B$ lies in the third quadrant and $\cos A = -\frac{\sqrt{3}}{2}$ and $\sin B = -\frac{3}{5}$,then find the value of $\frac{2 \tan B + \sqrt{3} \tan A}{\cot^2 A + \cos B}$.

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