If cotα = 1 and secβ = - (5)/(3) , where π < α < (3π)/(2) and...

Given,

$cot\alpha = 1, sec\beta = \dfrac{- 5}{3}$

So, $cos\beta = \dfrac{- 3}{5}, tan\beta = \dfrac{- 4}{3}$ and $tan\alpha = 1$

Now using formula $\tan \left(\alpha + \beta\right) = \dfrac{tan\alpha + tan\beta}{1 - tan\alpha tan\beta}$

$⇒ \tan \left(\alpha + \beta\right) = \dfrac{1 - \dfrac{4}{3}}{1 + \dfrac{4}{3} \times 1} = \dfrac{- 1}{7}$

We know that $tan\theta$ is negative in ${2}^{nd} \text{ and } {4}^{th}$ quadrant but as $\pi < \alpha < \dfrac{3\pi}{2}$, so $\alpha + \beta$ will lie in ${\text{IV}}^{\text{th}}$ quadrant.