A) \[\sqrt{2}\cos \theta ,\,\,\sqrt{2}\sin \theta \]
B) \[\sqrt{2}\sin \theta ,\,\,\sqrt{2}\cos \theta \]
C) \[\sqrt{2}\sin \theta ,\,\,\sqrt{2}\sin \theta \]
D) \[\sqrt{2}\,\cos \theta ,\,\,\sqrt{2}\,\cos \theta \]
Correct Answer: A
Solution :
We have \[\tan \theta =\frac{\sin \alpha -\cos \alpha }{\sin \alpha +\cos \alpha }\] \[\Rightarrow \tan \theta =\frac{\sin \left( \alpha -\frac{\pi }{4} \right)}{\cos \left( \alpha -\frac{\pi }{4} \right)}\Rightarrow \tan \theta =\tan \left( \alpha -\frac{\pi }{4} \right)\] \[\Rightarrow \theta =\alpha -\frac{\pi }{4}\Rightarrow \alpha =\theta +\frac{\pi }{4}\] Hence, \[\sin \alpha +\cos \alpha =\sin \left( \theta +\frac{\pi }{4} \right)+\cos \left( \theta +\frac{\pi }{4} \right)\] \[=\sqrt{2}\cos \theta \] and \[\sin \alpha -\cos \alpha =\sin \left( \theta +\frac{\pi }{4} \right)-\cos \left( \theta +\frac{\pi }{4} \right)\] \[=\frac{1}{\sqrt{2}}\sin \theta +\frac{1}{\sqrt{2}}\cos \theta -\frac{1}{\sqrt{2}}\cos \theta +\frac{1}{\sqrt{2}}\sin \theta \] \[=\frac{2}{\sqrt{2}}\sin \theta =\sqrt{2}\sin \theta \].
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