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Question: The value of \(\theta \) for which\(\dfrac{{2 + 3i\sin \theta }}{{1 - 2i\sin \theta }}\) is purely i...

The value of θ\theta for which2+3isin⁡θ1−2isin⁡θ\dfrac{{2 + 3i\sin \theta }}{{1 - 2i\sin \theta }} is purely imaginary is
A. π3\dfrac{\pi }{3}
B. π6\dfrac{\pi }{6}
C. sin⁡−1(34){\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{4}} \right)
D. sin⁡−1(13){\sin ^{ - 1}}\left( {\dfrac{1}{{\sqrt 3 }}} \right)

Explanation

Solution

In this question, first we will simplify the expression by making the denominator a real number by rationalization. After this we will separate the real and imaginary terms and then equate the real terms to zero. Finally solve the trigonometric equation to get the answer.

Complete step-by-step answer:
We have to find the value of if the given complex number2+3isin⁡θ1−2isin⁡θ\dfrac{{2 + 3i\sin \theta }}{{1 - 2i\sin \theta }} is purely imaginary.
For the given expression to be purely imaginary, its real part must be zero.
We will first simplify the expression and make the denominator a real number.
So we will rationalize the denominator part.
2+3isin⁡θ1−2isin⁡θ×1+2isin⁡θ1+2isin⁡θ\dfrac{{2 + 3i\sin \theta }}{{1 - 2i\sin \theta }} \times \dfrac{{1 + 2i\sin \theta }}{{1 + 2i\sin \theta }}
Now let’s simply the numerator and denominator part:
(2+3isin⁡θ)(1+2isin⁡θ)(1−2isin⁡θ)(1+2isin⁡θ)=2+7isin⁡θ−6sin⁡2θ(1−2isin⁡θ)(1+2isin⁡θ)\dfrac{{(2 + 3i\sin \theta )(1 + 2i\sin \theta )}}{{(1 - 2i\sin \theta )(1 + 2i\sin \theta )}} = \dfrac{{2 + 7i\sin \theta - 6{{\sin }^2}\theta }}{{(1 - 2i\sin \theta )(1 + 2i\sin \theta )}}
We know that a2−b2=(a+b)(a−b){a^2} - {b^2} = (a + b)(a - b) and also i2=−1{i^2} = - 1
∴\therefore The above expression can be written as:
2+7isin⁡θ−6sin⁡2θ12−(2isin⁡θ)2=2+7isin⁡θ−6sin⁡2θ1+4sin⁡2θ\dfrac{{2 + 7i\sin \theta - 6{{\sin }^2}\theta }}{{{1^2} - {{(2i\sin \theta )}^2}}} = \dfrac{{2 + 7i\sin \theta - 6{{\sin }^2}\theta }}{{1 + 4{{\sin }^2}\theta }}.
On separating the real part and imaginary part, we have:
2+7isin⁡θ−6sin⁡2θ1+4sin⁡2θ=2−6sin⁡2θ1+4sin⁡2+i7sin⁡θ1+4sin⁡2\dfrac{{2 + 7i\sin \theta - 6{{\sin }^2}\theta }}{{1 + 4{{\sin }^2}\theta }} = \dfrac{{2 - 6{{\sin }^2}\theta }}{{1 + 4{{\sin }^2}}} + i\dfrac{{7\sin \theta }}{{1 + 4{{\sin }^2}}}
On equating the real part to zero, we get:
2−6sin⁡2θ1+4sin⁡2\dfrac{{2 - 6{{\sin }^2}\theta }}{{1 + 4{{\sin }^2}}}=0
⇒2−6sin⁡2θ=0\Rightarrow 2 - 6{\sin ^2}\theta = 0
On solving, we get:
⇒sin⁡2θ=26\Rightarrow {\sin ^2}\theta = \dfrac{2}{6}
On taking square root on both sides, we have:
⇒sin⁡θ=±13\Rightarrow \sin \theta = \pm \sqrt {\dfrac{1}{3}} .
⇒θ=sin⁡−1(13) or sin⁡−1(−13)\Rightarrow \theta = {\sin ^{ - 1}}\left( {\dfrac{1}{{\sqrt 3 }}} \right){\text{ or }}{\sin ^{ - 1}}\left( { - \dfrac{1}{{\sqrt 3 }}} \right)

So, the correct answer is “Option D”.

Note: In such a type of question to make purely imaginary means we have to make the real part zero and if it asks for purely real then we will have to make the imaginary part zero. The purely imaginary term lies on the jωj\omega axis or imaginary axis of the complex plane and the purely real term lies on the real axis.