Let A = { [ 0 , 2 ] : 1 + 10 Re ( 2 cos + i sin cos – 3 i sin ) = 0 } . Then is equal to [2025]

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Solution

Q.
Let $A = {θ \in [0, 2 π] : 1 + 10 Re (\frac{2 \cos θ + i \sin θ}{\cos θ - 3 i \sin θ}) = 0}$ .

Then $\sum_{θ \in A} θ^{2}$ is equal to [2025]

1 $\frac{27}{4} π^{2}$

2 $6 π^{2}$

3 $\frac{21}{4} π^{2}$

4 $8 π^{2}$

Ans.
(3)

We have,

$\frac{2 \cos θ + i \sin θ}{\cos θ - 3 i \sin θ} = \frac{2 \cos θ + i \sin θ}{\cos θ - 3 i \sin θ} \times \frac{\cos θ + 3 i \sin θ}{\cos θ + 3 i \sin θ}$

$= \frac{2 \cos^{2} θ + 6 i \cos θ \sin θ + i \sin θ \cos θ - 3 \sin^{2} θ}{\cos^{2} θ + 9 \sin^{2} θ}$

$= \frac{2 \cos^{2} θ - 3 \sin^{2} θ}{\cos^{2} θ + 9 \sin^{2} θ} + \frac{i (7 \cos θ \sin θ)}{\cos^{2} θ + 9 \sin^{2} θ}$

Now, $1 + 10 Re (\frac{2 \cos θ + i \sin θ}{\cos θ - 3 i \sin θ}) = 0$

$\Rightarrow 1 + \frac{20 \cos^{2} θ - 30 \sin^{2} θ}{\cos^{2} θ + 9 \sin^{2} θ} = 0$

$\Rightarrow \cos^{2} θ = \sin^{2} θ$

$\Rightarrow \tan^{2} θ = 1 \Rightarrow \tan θ = \pm 1$

$\Rightarrow θ = \frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4}$

$∴ \sum_{θ \in A} θ^{2} = \frac{π^{2}}{16} + \frac{9 π^{2}}{16} + \frac{25 π^{2}}{16} + \frac{49 π^{2}}{16} = \frac{84 π^{2}}{16} = \frac{21 π^{2}}{4}$

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