JEE Mains · Maths · STD 11 - 4.1 complex nubers
If the real part of the complex number \(z=\frac{3+2 i \cos \theta}{1-3 i \cos \theta}, \theta \in\left(0, \frac{\pi}{2}\right)\) is zero, then the value of \(\sin ^{2} 3 \theta+\cos ^{2} \theta\) is equal to \(.....\)
- A \(1\)
- B \(3\)
- C \(2\)
- D \(4\)
Answer & Solution
Correct Answer
(A) \(1\)
Step-by-step Solution
Detailed explanation
\(\operatorname{Re}(z)=\frac{3-6 \cos ^{2} \theta}{1+9 \cos ^{2} \theta}=0\) \(\Rightarrow \theta=\frac{\pi}{4}\) Hence, \(\sin ^{2} 3 \theta+\cos ^{2} \theta=1\)
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