A value of θ for which 2+3isinθ/(1-isinθ) is purely imaginary is: π/3, π/6, sin⁻¹(√3/4), sin⁻¹(1/√3)

⇒2+3isinθ/(1-isinθ)
Purely imaginary means real part=0
2+3isinθ/(1-isinθ) × (1+2isinθ)/(1+2isinθ) = (2+4isinθ+3isinθ-6sin²θ)/(1+4sin²θ) = (2+7isinθ-6sin²θ)/(1+4sin²θ)
Real part is = (2-6sin²θ)/(1+4sin²θ)
This is equal to zero. ⇒(2-6sin²θ)/(1+4sin²θ)=0
So, 2-6sin²θ=0
2=6sin²θ
sin²θ=1/3
sinθ=±1/√3
∴ θ=sin⁻¹(1/√3)