A value of θ for which \frac{2\: \: \: \: \: 3i\: \sin \theta }{1-2i\sin \theta }  is purely imaginary, is :

 

  • Option 1)

    \frac{\pi }{3}

  • Option 2)

    \frac{\pi }{6}

  • Option 3)

    \sin ^{-1}\left ( \frac{\sqrt{3}}{4} \right )

  • Option 4)

    \sin ^{-1}\left ( \frac{1}{\sqrt{3}} \right )

 

Answers (2)

As we learnt in

Purely Real Complex Number -

z=x+iy, x\epsilon R, \boldsymbol{y=0}

& i2=-1

- wherein

Real part of z = Re (z) = x & Imaginary part of z = Im (z) = y

 

 

Let  Z=\frac{2+3i\ sin\theta}{1-2i\ sin\theta}

Z=\frac{2+3i\ sin\theta}{1-2i\ sin\theta}\times \frac{1+2i\ sin\theta}{1+2i\ sin\theta}

    =\frac{(2+3i\ sin\theta)(1+2i\ sin\theta)}{1+4sin^{2}\theta}

    =2-6sin^{2}\theta=0            for pure for real part must be zero.

    sin^{2}\theta=\frac{2}{6}=\frac{1}{3}

    sin^{2}\theta=\frac{1}{\sqrt{3}}

    \theta=sin^{-1}\frac{1}{\sqrt{3}}

Correct option is 4.

 


Option 1)

\frac{\pi }{3}

This is an incorrect option.

Option 2)

\frac{\pi }{6}

This is an incorrect option.

Option 3)

\sin ^{-1}\left ( \frac{\sqrt{3}}{4} \right )

This is an incorrect option.

Option 4)

\sin ^{-1}\left ( \frac{1}{\sqrt{3}} \right )

This is the correct option.

N neha

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