Let A=θ∈−π2,π:3+2isin⁡θ1−2isin⁡θ is purely imaginary . Then the sum of the elements in A .is

# Let . Then the sum of the elements in A .is

1. A

$3\pi /4$

2. B

$2\pi /3$

3. C

$\pi$

4. D

$5\pi /6$

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### Solution:

Let $z=\frac{3+2i\mathrm{sin}\theta }{1-2i\mathrm{sin}\theta }$

$=\frac{3+2i\mathrm{sin}\theta }{1-2i\mathrm{sin}\theta }×\frac{1+2i\mathrm{sin}\theta }{1+2i\mathrm{sin}\theta }=\frac{\left(3-4{\mathrm{sin}}^{2}\theta \right)+i\left(8\mathrm{sin}\theta \right)}{1+4{\mathrm{sin}}^{2}\theta }$

Now, $\left(z\right)=0$                                        [ $\because z$ is purely imaginary]

$\theta =-\frac{\pi }{3},\frac{\pi }{3},\frac{2\pi }{3}$ $\left[\because \theta \in \left(\frac{-\pi }{2},\pi \right)\right]$

Hence, sum of elements of $f\left(A\right)=\frac{-\pi }{3}+\frac{\pi }{3}+\frac{2\pi }{3}=\frac{2\pi }{3}$  Register to Get Free Mock Test and Study Material

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