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STD 11 - 3. trignometrical ratios,functions and identities — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 3. trignometrical ratios,functions and identities4 Marks
MCQ
Given that $\pi < \alpha < \frac{{3\pi }}{2},$ then the expression $\sqrt {(4{{\sin }^4}\alpha + {{\sin }^2}2\alpha )} + 4{\cos ^2}\left( {\frac{\pi }{4} - \frac{\alpha }{2}} \right)$ is equal to
  • A
    $2$
  • B
    $2 - 4\sin \alpha $
  • ✓
    $(a)$ and $(b)$
  • D
    None of these
✓

Answer

Correct option: C.
$(a)$ and $(b)$
c
(c) 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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