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 \[u=\cos \theta \left\{ \sin \theta +\sqrt{{{\sin }^{2}}\theta +{{\sin }^{2}}\alpha } \right\}\] Þ \[{{(u-\sin \theta \cos \theta )}^{2}}={{\cos }^{2}}\theta ({{\sin }^{2}}\theta +{{\sin }^{2}}\alpha )\] Þ \[{{u}^{2}}{{\tan }^{2}}\theta -2u\tan \theta +{{u}^{2}}-{{\sin }^{2}}\alpha =0\] Since tan \[\theta \]is real, therefore Þ \[4{{u}^{2}}-4{{u}^{2}}({{u}^{2}}-{{\sin }^{2}}\alpha )\ge 0\] \[\Rightarrow {{u}^{2}}-(1+{{\sin }^{2}}\alpha )\le 0\] Þ \[|u|\,\le \sqrt{1+{{\sin }^{2}}\alpha }\].
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