A) \[\sqrt{3}\,\cos \theta \]
B) \[\sqrt{2}\,\sin \theta \]
C) \[\sqrt{2}\,\cos \theta \]
D) \[\sqrt{3}\,\sin \theta \]
Correct Answer: D
Solution :
| [d] \[\sin \theta +\cos \theta =\sqrt{2}\,\cos \theta \] \[\Rightarrow \] \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta +2\sin \theta \cos \theta =2{{\cos }^{2}}\] \[\Rightarrow \] \[1-\sin 2\theta =2{{\cos }^{2}}\theta \] \[\Rightarrow \] \[\sin 2\theta =2{{\cos }^{2}}\theta \] \[\Rightarrow \] \[\sin 2\theta =\cos 2\theta \] Now, \[(co{{s}^{2}}\theta +si{{n}^{2}}\theta )=\sqrt{{{(cos\theta -sin\theta )}^{2}}}\] \[=\sqrt{{{\cos }^{2}}\theta +{{\sin }^{2}}\theta -2\cos \theta \cdot \sin \theta }\] \[=\sqrt{1-\sin 2\theta }=\sqrt{1-\cos 2\theta }\] \[=\sqrt{2{{\sin }^{2}}\theta }=\sqrt{2}\sin \theta \] |
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