If `(3+2i sintheta)/(1-2i sintheta)` is a real number and `0lt theta lt 2pi`, then `theta` is `a. pi \ b. pi/2 \ c. pi/3 \ d. pi/6`

Explanation

`frac(3+2 i sin theta)(1-2 i sin theta)` is a real number

On rationalising, we get,



`frac(3+2 i sin theta)(1-2 i sin theta) times frac(1+2 i sin theta)(1+2 i sin theta )`

`=frac((3+2 i sin theta)(1+2 i sin theta))((1)^2-(2 i sin theta)^2)`

`=frac(3+2 i sin theta+6 i sin theta+4 i^2 sin ^2 theta)(1+4 sin ^2 theta)`

`=frac(3-4 sin ^2 theta+8 i sin theta)(1+4 sin ^2 theta)[i^2=-1]`

`=frac(3-4 sin ^2 theta)(1+4 sin ^2 theta)+i frac(8 sin theta)(1+4 sin ^2 theta) `
For the above term to be real, the imaginary part has to be zero.
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