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# If $\theta$ lies in the third quadrant, then the expression $\sqrt {4{{\sin }^4}\theta + {{\sin }^2}2\theta } + 4{\cos ^2}(\dfrac{\pi }{4} - \dfrac{\theta }{2})$ equals 2.A. TrueB. False

Last updated date: 21st Jun 2024
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Hint: Here we will solve the given equation using the trigonometric identities and check whether the value of expression equals 2 or not.

As we know that $\sin 2\theta = 2\sin \theta \cos \theta \to (1)$
$\Rightarrow \sqrt {4{{\sin }^4}\theta + {{\sin }^2}2\theta } + 4{\cos ^2}(\dfrac{\pi }{4} - \dfrac{\theta }{2}) \to (2)$
Putting the value of $\sin 2\theta$ as in equation (1) in equation (2) then we have
$\Rightarrow \sqrt {4{{\sin }^4}\theta + {{\sin }^2}2\theta } = \sqrt {4{{\sin }^2}\theta ({{\sin }^2}\theta + {{\cos }^2}\theta )} = \sqrt {4{{\sin }^2}\theta } = 2\left| {\sin \theta } \right|.$
Now as per question, $\theta$ lies in the third quadrant, so $\sin \theta < 0$ and $\left| {\sin \theta } \right| = - \sin \theta$
Hence $\sqrt {4{{\sin }^4}\theta + {{\sin }^2}2\theta } = - 2\sin \theta$
And $4{\cos ^2}(\dfrac{\pi }{4} - \dfrac{\theta }{2}) = 2[1 + \cos \left( {\dfrac{\pi }{2} - \theta } \right)] = 2 + 2\sin \theta {\text{ [}}\because {\text{cos2}}\theta {\text{ = 2co}}{{\text{s}}^2}\theta - 1] \to (3)$
$\Rightarrow - 2\sin \theta + 2 + 2\sin \theta = 2$
Note: Equation (1) and (3) is trigonometry identity, while solving any question always try to expand the expression by putting the value of trigonometry identities. Remember $\left| x \right| = x$ if $x \geqslant 0$ and $\left| x \right| = - x$ if $x \leqslant 0$. In the third quadrant only $\tan \theta$ and $\cot \theta$ is positive.