If cot alpha = 1 and sec beta = -frac{5}{3},w | vedclass

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(A) Given $\cot \alpha = 1$. Since $\pi < \alpha < \frac{3\pi}{2}$ (third quadrant),$	an \alpha = 1$.Given $\sec \beta = -\frac{5}{3}$. Since $\frac{

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$-\frac{1}{7}$ and $IV^{th}$ quadrant

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(A) Given $\cot \alpha = 1$. Since $\pi Given $\sec \beta = -\frac{5}{3}$. Since $\frac{\pi}{2} Using $	an^2 \beta = \sec^2 \beta - 1$,we get $	an^2 \beta = (-\frac{5}{3})^2 - 1 = \frac{25}{9} - 1 = \frac{16}{9}$.Since $\beta$ is in the second quadrant,$	an \beta = -\frac{4}{3}$.Now,$	an(\alpha + \beta) = \frac{	an \alpha + 	an \beta}{1 - 	an \alpha 	an \beta} = \frac{1 + (-4/3)}{1 - (1)(-4/3)} = \frac{-1/3}{1 + 4/3} = \frac{-1/3}{7/3} = -\frac{1}{7}$.Since $	an(\alpha + \beta) = -\frac{1}{7} Given $\pi This range corresponds to the fourth quadrant ($270^\circ$ to $360^\circ$) or the first quadrant ($360^\circ$ to $450^\circ$).Thus,$\alpha + \beta$ lies in the $IV^{th}$ quadrant.

If $\cot \alpha = 1$ and $\sec \beta = -\frac{5}{3}$,where $\pi < \alpha < \frac{3\pi}{2}$ and $\frac{\pi}{2} < \beta < \pi$,then the value of $\tan(\alpha + \beta)$ and the quadrant in which $\alpha + \beta$ lies,respectively,are

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