A value of θ for which 1−2isinθ2+3isinθ is purely imaginary, is:
(A)6π
(B)sin−1(43)
(C)sin−1(31)✓ Correct
(D)3π
Answer: (C)
Solution
Multiply top and bottom by the conjugate of the denominator: 1+4sin2θ(2+3isinθ)(1+2isinθ)=1+4sin2θ(2−6sin2θ)+7isinθ. The number is purely imaginary when its real part vanishes, 2−6sin2θ=0, so sin2θ=31 and sinθ=31. A value of θ is therefore sin−1(31).
JEE Main · 2016 · 03 Apr 2016 · Offline · Single correct option