If A lies in the third quadrant and 3tanA - 4 = 0 then 5sin2A + 3sinA + 4cosA is equal to (1) 0 (2) -24/5 (3) 24/5 (4) 48/5

 Answer: (1) 0

3tanA – 4 = 0

3tanA = 4

tanA = 4/3

tan2A = 16/9

sec2A = 1 + tan2A

sec2A = 1 + 16/9 = 25/9

Since A lies in third quadrant

sec A = -5/3 and cos A = -3/5

sin2A = 1−cos2A

sin2A = 1 – 9/25

sin2A = 16/25

sin A = -4/5

Need to find: 5sin2A + 3sinA + 4cosA

sin2A = 2 sinA cosA

5[2 sinA cosA] + 3sinA + 4cosA = 5 [2 × (-4/5) × (-3/5)] + 3 × (-4/5) + 4 ×(-3/5)

5[2 sinA cosA] + 3sinA + 4cosA = 10 × 12/15 – 512 – 12/ 5

5sin2A + 3sinA + 4cosA = 0