If cot α = 1 and sec β = - 5 3 , where π < α < 3 π 2 and π 2 < β < π , then the value of tan α + β and the quadrant in which α + β lies, respectively are

Step-by-step solution

Given, cot α = 1 , sec β = - 5 3 So, cos β = - 3 5 , tan β = - 4 3 and tan α = 1 Now using formula tan α + β = tan α + tan β 1 - tan α tan β ⇒ tan α + β = 1 - 4 3 1 + 4 3 × 1 = - 1 7 We know that tan θ is negative in 2 n d   &   4 t h quadrant but as π < α < 3 π 2 , so α + β will lie in IV th quadrant.