Here, A, B, C are in AP, so 2B = A + C = 30 + C .....(1) ABC is a triangle, so A + B + C =180° ⇒ B = 150 - C .....(2) Add (1) and (2), we get 3B = 180 ⇒ B = 60° So, C = 90° 2sinA + 3tanB - 4cosC = 2(sin 30) + 3 tan (60) - 4 cos 90 =2(1∕2)+3√3 =1+3√3 Hence, option (4) is correct.