Question

Let 0 < A, B < π/2 satisfying the equation 3sin² A + 2sin² B = 1 and 3sin 2A – 2sin 2B = 0, then A + 2B is equal to ________

(a) π

(b) π/2

(c) π/4

(d) 2π

Answer 1

Correct option is (b) π/2 

3 sin 2A – 2sin 2B = 0 

sin 2B = 3/2 sin 2A …….(i) 

3 sin² A + 2 sin² B = 1 

3 sin² A = 1 – 2 sin² B 

3 sin² A = cos 2B ……(ii) 

cos(A + 2B) = cos A cos 2B – sin A sin 2B 

= cos A (3 sin² A) – sin A (3/2 sin 2A) …..[From (i) and (ii)] 

= 3 cos A sin² A – 3/2 (sin A) (2 sin A cos A) 

= 3 cos A sin² A – 3 sin² A cos A 

= 0

= cos π/2

∴ A + 2B = π/2 ……..[∵ 0 < A + 2B < 3π/2]