A value of θ for which 2+3isinθ1−2isinθ is purely imaginary, is:
A
sin−1(√34)
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B
sin−1(1√3)
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C
π3
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D
π6
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Solution
The correct option is Bsin−1(1√3) Rationalizing the given expression (2+3isinθ)(1+2isinθ)1+4sin2θ For the given expression to be purely imaginary, real part of the above expression should be equal to zero. ⇒2−6sin2θ1+4sin2θ=0⇒sin2θ=13⇒sinθ=±1√3