Question

Let 0 < A, B < π/2 satisfying the equalities 3 sin² A + 2 sin² B = 1 and 3sin2A – 2sin2B = 0. Then A + 2B =

(A) π/4

(B)  π/3

(C)  π/2

(D)  None of these

Answer 1

Correct option (C)  π/2

From the second equation, we have

sin2B = 3/2sin2A

and from the first equality

3sin² A = 1 –2 sin2B = cos2B

Now cos (A + 2B) = cosA⋅cos2B – sinA⋅sin2B

= 3 cosA⋅sin²A - 3/2 ⋅sinA⋅sin2A

= 3cosA⋅sin2A – 3sin2A⋅cosA = 0

⇒ A + 2B = π/2 or 3π/2

Given that 0 < A π/2 and 0 < B < π/2

⇒ 0 < A + 2B < π + π/2

Hence, A + 2B = π/2