Let 0 < A, B < π/2 satisfying the equation 3sin2 A + 2sin2 B = 1 and

1.	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» 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]

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»

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]