A value of θ for which 2 + 3 i sin θ 1 - 2 i sin θ is purely imaginary, is

Step-by-step solution

Let, z = 2  +  3 i sin   θ 1  -  2 i sin   θ Multiplying the denominator by its conjugate, we get, z = 2 + 3   i sin   θ   1 + 2   i sin   θ 1 + 4   sin 2   θ = 2 - 6  sin 2   θ + i 7 sin   θ 1 + 4  sin 2   θ To make it purely imaginary. Its real part should be 0 . Hence, 2 = 6  sin 2 ⁡ θ ⇒ sin ⁡ θ = ± 1 3