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

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

$\frac{1}{7}$ and I^st quadrant

- 7 and IV^th quadrant

7 and I^st quadrant

answer is A.

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Detailed Solution

$\begin{array}{l} \cot α = 1, \sec β = \frac{- 5}{3}, \cos β = \frac{- 3}{5}, \tan β = \frac{- 4}{3} \\ \tan (α + β) = \frac{1 - \frac{4}{3}}{1 + \frac{4}{3} \times 1} = \frac{- 1}{7} < 0 \\ π < α < \frac{3 π}{2} and \frac{π}{2} < β < π \\ \Rightarrow \frac{3 π}{2} < α + β < \frac{5 π}{2} \Rightarrow α + β \in IV Quadrant \end{array}$