If sin A = frac{3}{5},tan B = frac{1}{2},and | vedclass

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(D) Given $\sin A = \frac{3}{5}$. Since $\frac{\pi}{2} < A < \pi$,$A$ lies in the second quadrant where $	an A$ is negative.Using $\cos^2 A = 1 - \si

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$-\frac{7}{2}$

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(D) Given $\sin A = \frac{3}{5}$. Since $\frac{\pi}{2} Using $\cos^2 A = 1 - \sin^2 A = 1 - (\frac{3}{5})^2 = 1 - \frac{9}{25} = \frac{16}{25}$,we get $\cos A = -\frac{4}{5}$.Thus,$	an A = \frac{\sin A}{\cos A} = \frac{3/5}{-4/5} = -\frac{3}{4}$.Given $	an B = \frac{1}{2}$. Since $\pi Using $\sec^2 B = 1 + 	an^2 B = 1 + (\frac{1}{2})^2 = 1 + \frac{1}{4} = \frac{5}{4}$,we get $\sec B = -\frac{\sqrt{5}}{2}$.Now,substitute these values into the expression $8 	an A - \sqrt{5} \sec B$:$= 8(-\frac{3}{4}) - \sqrt{5}(-\frac{\sqrt{5}}{2})$$= -6 + \frac{5}{2} = \frac{-12 + 5}{2} = -\frac{7}{2}$.

If $\sin A = \frac{3}{5}$,$\tan B = \frac{1}{2}$,and $\frac{\pi}{2} < A < \pi < B < \frac{3\pi}{2}$,then the value of $8 \tan A - \sqrt{5} \sec B$ is:

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