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JEE Main 2016

JEE Main 2016 Mathematics question on Complex Numbers and Quadratic Equations

A value of θ\theta for which 2+3isin⁡θ1−2isin⁡θ\dfrac{2 + 3i\sin\theta}{1 - 2i\sin\theta} is purely imaginary, is
  1. A.
    π3\dfrac{\pi}{3}
  2. B.
    π6\dfrac{\pi}{6}
  3. C.
    sin⁡−1(34)\sin^{-1}\left(\dfrac{\sqrt{3}}{4}\right)
  4. D.
    sin⁡−1(13)\sin^{-1}\left(\dfrac{1}{\sqrt{3}}\right)
Show answer and solution

Correct option: D

Solution

Multiply numerator and denominator by the conjugate of the denominator (1 + 2i sinθ):

Numerator: (2 + 3i sinθ)(1 + 2i sinθ) = 2 + 4i sinθ + 3i sinθ + 6i²sin²θ = (2 - 6sin²θ) + 7i sinθ

For purely imaginary, the real part must be zero:
2 - 6sin²θ = 0 → sin²θ = 1/3 → sinθ = 1/√3 → θ = sin⁻¹(1/√3)

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