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If $\theta \in\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)$ then the value of $\sqrt{4 \cos ^{4} \theta+\sin ^{2} 2 \theta}+4 \cot \theta \cos ^{2}\left(\frac{\pi}{4}-\frac{\theta}{2}\right)$ is
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The correct answer is:
$2 \cot \theta$
$\sqrt{4 \cos ^{4} \theta+\sin ^{2} 2 \theta}+4 \cot \theta \cos ^{2}\left(\frac{\pi}{4}-\frac{\theta}{2}\right)$
$=\sqrt{4 \cos ^{4} \theta+(2 \sin \theta \cos \theta)^{2}} +2 \cot \theta\left[2 \cos ^{2}\left(\frac{\pi}{4}-\frac{\theta}{2}\right)\right]$
$=\sqrt{4 \cos ^{4} \theta+4 \sin ^{2} \theta \cos ^{2} \theta} +2 \cot \theta\left[1+\cos \left(\frac{\pi}{2}-\theta\right)\right]$
$=\sqrt{4 \cos ^{2} \theta\left(\cos ^{2} \theta+\sin ^{2} \theta\right)}+2 \cot \theta(1+\sin \theta)$
$=|2 \cos \theta|+2 \cot \theta+2 \cos \theta$
$=-2 \cos \theta+2 \cot \theta+2 \cos \theta$
$\quad\left[\right.$ for $\left.\theta \in\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right),|\cos \theta|=-\cos \theta\right]$
$=2 \cot \theta$
$=\sqrt{4 \cos ^{4} \theta+(2 \sin \theta \cos \theta)^{2}} +2 \cot \theta\left[2 \cos ^{2}\left(\frac{\pi}{4}-\frac{\theta}{2}\right)\right]$
$=\sqrt{4 \cos ^{4} \theta+4 \sin ^{2} \theta \cos ^{2} \theta} +2 \cot \theta\left[1+\cos \left(\frac{\pi}{2}-\theta\right)\right]$
$=\sqrt{4 \cos ^{2} \theta\left(\cos ^{2} \theta+\sin ^{2} \theta\right)}+2 \cot \theta(1+\sin \theta)$
$=|2 \cos \theta|+2 \cot \theta+2 \cos \theta$
$=-2 \cos \theta+2 \cot \theta+2 \cos \theta$
$\quad\left[\right.$ for $\left.\theta \in\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right),|\cos \theta|=-\cos \theta\right]$
$=2 \cot \theta$
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