A) \[\sqrt{1+{{\cos }^{2}}\alpha }\]
B) \[\sqrt{1+{{\sin }^{2}}\alpha }\]
C) \[\sqrt{2+{{\sin }^{2}}\alpha }\]
D) \[\sqrt{2+{{\cos }^{2}}\alpha }\]
Correct Answer: B
Solution :
Let\[u=\cos \theta \{\sin \theta +\sqrt{{{\sin }^{2}}\theta +{{\sin }^{2}}\alpha }\}\] \[\Rightarrow \]\[{{(u-\sin \theta \cdot \cos \theta )}^{2}}={{\cos }^{2}}\theta \{{{\sin }^{2}}\theta +{{\sin }^{2}}\alpha \}\] \[\Rightarrow \]\[{{x}^{2}}{{\tan }^{2}}\theta -2u\tan \theta +{{u}^{2}}-{{\sin }^{2}}\alpha =0\] Since, \[\tan \theta \] is real, therefore \[4{{u}^{2}}-4{{u}^{2}}({{u}^{2}}-{{\sin }^{2}}\alpha )\ge 0\] \[\Rightarrow \] \[{{u}^{2}}-(1+{{\sin }^{2}}\alpha )\ge 0\] \[\Rightarrow \] \[|u|\le \sqrt{1+{{\sin }^{2}}\alpha }\]
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