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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Trignometry
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If $\pi < \theta < \large\frac{3\pi}{2}$ the expression $\sqrt{4\sin^4\theta+\sin^22\theta}+4\cos^2\big(\large\frac{\pi}{4}-\frac{\theta}{2}\big)$ is equal to

$\begin{array}{1 1}(a)\;2&(b)\;2+4\sin\theta\\(c)\;2-4\sin\theta&(d)\;None\;of\;these\end{array}$

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1 Answer

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Given expression :
$\sqrt{4\sin^4\theta+\sin^2\theta}+4\cos^2\big(\large\frac{\pi}{2}-\frac{\theta}{2}\big)$
$\Rightarrow \sqrt{4\sin^4\theta+4\sin^2\theta\cos^2\theta}+2.2\cos^2\big(\large\frac{\pi}{2}-\frac{\theta}{2}\big)$
$\Rightarrow \sqrt{4\sin^2\theta(\sin^2\theta+\cos^2\theta)}+2\big[1+\cos 2\big(\large\frac{\pi}{2}-\frac{\theta}{2}\big)\big]$
$\Rightarrow 2\sqrt{\sin^2\theta}+2+2\cos\big(\large\frac{\pi}{2}$$-\theta\big)$
$\Rightarrow 2\mid \sin\theta\mid+2\sin\theta+2$
$\sqrt{z^2}=\mid z\mid$
$\Rightarrow -2\sin\theta+2\sin\theta+2$
$\pi < \theta <\large\frac{3\pi}{2}$
$\Rightarrow \sin\theta$ is -ve
$\Rightarrow 2$
Hence (a) is the correct answer.
answered Oct 17, 2013 by sreemathi.v
 

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