A) 2
B) \[2+4\sin \alpha \]
C) \[2-4\sin \alpha \]
D) None of these
Correct Answer: C
Solution :
Given that \[\pi <\alpha <\frac{3\pi }{2}i.e.,\alpha \]is in third quadrant. Now, \[\sqrt{(4{{\sin }^{4}}\alpha +{{\sin }^{2}}2\alpha )}+4{{\cos }^{2}}\left( \frac{\pi }{4}-\frac{\alpha }{2} \right)\] \[=\sqrt{(4{{\sin }^{4}}\alpha +4{{\sin }^{2}}\alpha {{\cos }^{2}}\alpha )}+2.2{{\cos }^{2}}\left( \frac{\pi }{4}-\frac{\alpha }{2} \right)\] \[=\sqrt{4{{\sin }^{2}}\alpha ({{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha )}+2\left[ 1+\cos \left( \frac{\pi }{2}-\alpha \right) \right]\] \[=\pm 2\sin \alpha +2+2\sin \alpha \] On taking -ve, answer is 2 and on taking +ve, answer is \[2+4\sin \alpha \] But\[\pi <\alpha <\frac{3\pi }{4},\]Hence answer is \[2-4\sin \alpha \]because \[\sin \alpha \]is \[-ve\]in third quadrant.
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