Let, z=1−2isinθ2+3isinθ
Multiplying the denominator by its conjugate, we get,
z=1+4sin2θ(2+3isinθ)(1+2isinθ)
=1+4sin2θ(2−6sin2θ)+i(7sinθ)
To make it purely imaginary. Its real part should be 0.
Hence, 2=6sin2θ
⇒sinθ=±31
JEE Main 2016 — Mathematics Algebra
A value of θ for which 1−2isinθ2+3isinθ is purely imaginary, is
Held on 3 Apr 2016 · Verified 6 Jul 2026.
sin−1(43)
sin−1(31)
3π
6π
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