A) \[2\] and \[-1\]
B) \[1\]and \[\frac{1}{2}\]
C) \[\frac{1}{2}\]and \[\frac{1}{3}\]
D) \[3\] and \[-4\]
Correct Answer: A
Solution :
| [a] \[\sqrt{-4+\sqrt{8+16\cos e{{c}^{4}}\theta +{{\sin }^{4}}\theta }}\] |
| \[=\sqrt{-4+\sqrt{8+\frac{16}{{{\sin }^{4}}\theta }+{{\sin }^{4}}\theta }}\]] |
| \[=\sqrt{-4+\sqrt{{{\left( \frac{4}{{{\sin }^{2}}\theta }+{{\sin }^{2}}\theta \right)}^{2}}}}\] |
| \[[{{a}^{2}}+2ab+{{b}^{2}}={{(a+b)}^{2}}]\] |
| \[=\sqrt{-4+\frac{4}{{{\sin }^{2}}\theta }+{{\sin }^{2}}\theta }\]\[[{{a}^{2}}-2ab+{{b}^{2}}={{(a-b)}^{2}}]\] |
| \[=\sqrt{{{\left( -\sin \theta +\frac{2}{\sin \theta } \right)}^{2}}}=2\cos ec\,\theta (-1)\sin \theta \] |
| \[=A\,\,\cos ec\,\,\theta +B\sin \theta \] |
| On comparing, we get |
| \[A=2\] and \[B=-1\] |
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