A) \[\sqrt{2}\cos \,\theta \]
B) \[\sqrt{2}sin\,\theta \]
C) \[0\]
D) \[1\]
Correct Answer: A
Solution :
Given, \[\cos \theta -\sin \theta =\sqrt{2}\,\sin \theta \] Squaring, we get \[{{\cos }^{2}}\theta +{{\sin }^{2}}\theta -2\sin \theta \,\cos \theta =2{{\sin }^{2}}\theta \] or \[1-2\sin \theta \,\cos \theta =2(1-{{\cos }^{2}}\theta )\] or \[1+2\sin \theta \,\cos \theta =2{{\cos }^{2}}\theta \] or \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta +2\sin \theta \,\cos \theta =2\,{{\cos }^{2}}\theta \] or \[{{(\sin \theta +\cos \theta )}^{2}}=2{{\cos }^{2}}\theta \] \[\therefore \] \[\sin \theta +\cos \theta =\sqrt{2}\,\cos \theta \]
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