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# If $\pi < \theta < \dfrac{{3\pi }}{2}$ the expression $\sqrt {4{{\sin }^4}\theta + {{\sin }^2}2\theta } + 4{\cos ^2}\left( {\dfrac{\pi }{4} - \dfrac{\theta }{2}} \right)$ is equal to(A) $2$ (B) $2 + 4\sin \theta$ (C) $2 - 4\sin \theta$ (D) $0$

Hint- Here in this question we will use some basic trigonometric identities as
$\sin 2\alpha = 2\sin \alpha \cos \alpha$
${\sin ^2}\alpha + {\cos ^2}\alpha = 1$
$1 + 2{\cos ^2}\alpha = \cos 2\alpha$
$\cos \left( {\dfrac{\pi }{2} - \theta } \right) = \sin \theta$

Complete step by step solution-
We have to simplify the expression$\sqrt {4{{\sin }^4}\theta + {{\sin }^2}2\theta } + 4{\cos ^2}\left( {\dfrac{\pi }{4} - \dfrac{\theta }{2}} \right)$; to do this we will use some basic identities here.
$\sqrt {4{{\sin }^4}\theta + {{\sin }^2}2\theta } + 4{\cos ^2}\left( {\dfrac{\pi }{4} - \dfrac{\theta }{2}} \right)$
$= \sqrt {4{{\sin }^4}\theta + {{\left( {2\sin \theta \cos \theta } \right)}^2}} + 4{\cos ^2}\left( {\dfrac{\pi }{4} - \dfrac{\theta }{2}} \right)$ [ ]
On opening the bracket,
$= \sqrt {4{{\sin }^4}\theta + \left( {4 \times {{\sin }^2}\theta \times {{\cos }^2}\theta } \right)} + 4{\cos ^2}\left( {\dfrac{\pi }{4} - \dfrac{\theta }{2}} \right)$
On taking the common term from the terms under the square root,
$= \sqrt {4{{\sin }^2}\theta \left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right)} + 4{\cos ^2}\left( {\dfrac{\pi }{4} - \dfrac{\theta }{2}} \right)$
$= \sqrt {4{{\sin }^2}\theta \times \left( 1 \right)} + 4{\cos ^2}\left( {\dfrac{\pi }{4} - \dfrac{\theta }{2}} \right)$ [ ]
On splitting the $4$ in to factors,
$= \sqrt {4{{\sin }^2}\theta } + 2 \times \left( {2{{\cos }^2}\left( {\dfrac{\pi }{4} - \dfrac{\theta }{2}} \right)} \right)$
$= \sqrt {4{{\sin }^2}\theta } + 2 \times \left( {\cos \left( {2 \times \left( {\dfrac{\pi }{4} - \dfrac{\theta }{2}} \right)} \right) + 1} \right)$ [ ]
$= 2\sin \theta + 2 \times \left( {\cos \left( {\dfrac{\pi }{2} - \theta } \right) + 1} \right)$
$= 2\sin \theta + 2 \times \left( {\sin \theta + 1} \right)$ []$= 2\sin \theta + 2\sin \theta + 2$
$= 4\sin \theta + 2$
Hence, $\sqrt {4{{\sin }^4}\theta + {{\sin }^2}2\theta } + 4{\cos ^2}\left( {\dfrac{\pi }{4} - \dfrac{\theta }{2}} \right) = 4\sin \theta + 2$
Therefore,Option B is the correct answer

Note: To solve this type of question we just need to learn all the identities such that we can find the particular form in the given expression so that an identity can be applied.Also, care has to be taken to apply the appropriate identity in accordance to the problem given