If cot alpha = frac{1}{2} and sec beta = -fra | vedclass

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(D) Given $\cot \alpha = \frac{1}{2}$,since $\alpha \in (\pi, \frac{3\pi}{2})$ (third quadrant),$	an \alpha = \frac{1}{\cot \alpha} = 2$. Given $\sec

Vedclass · Accepted Answer

$\frac{2}{11}$

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(D) Given $\cot \alpha = \frac{1}{2}$,since $\alpha \in (\pi, \frac{3\pi}{2})$ (third quadrant),$	an \alpha = \frac{1}{\cot \alpha} = 2$. Given $\sec \beta = -\frac{5}{3}$,since $\beta \in (\frac{\pi}{2}, \pi)$ (second quadrant),$	an^2 \beta = \sec^2 \beta - 1 = (-\frac{5}{3})^2 - 1 = \frac{25}{9} - 1 = \frac{16}{9}$. Since $\beta$ is in the second quadrant,$	an \beta$ must be negative,so $	an \beta = -\frac{4}{3}$. Using the formula $	an(\alpha + \beta) = \frac{	an \alpha + 	an \beta}{1 - 	an \alpha 	an \beta}$: $	an(\alpha + \beta) = \frac{2 + (-\frac{4}{3})}{1 - (2)(-\frac{4}{3})} = \frac{\frac{6-4}{3}}{1 + \frac{8}{3}} = \frac{\frac{2}{3}}{\frac{11}{3}} = \frac{2}{11}$.

If $\cot \alpha = \frac{1}{2}$ and $\sec \beta = -\frac{5}{3}$,where $\alpha \in \left(\pi, \frac{3\pi}{2}\right)$ and $\beta \in \left(\frac{\pi}{2}, \pi\right)$,then the value of $\tan(\alpha + \beta)$ is:

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