| If \[\cos \theta +\sin \theta =\sqrt{2}\cos \theta .\]Then, the value of \[\cos \theta -\sin \theta \]is |
A) \[\sqrt{3}\sin \theta \]
B) \[\sqrt{2}\cos \theta \]
C) \[\sqrt{2}\sin \theta \]
D) \[\sqrt{3}\cos \theta \]
Correct Answer: C
Solution :
| Given, \[\cos \theta +\sin \theta =\sqrt{2}\cos \theta \] |
| On squaring both sides, we get |
| \[{{(\cos \theta +\sin \theta )}^{2}}={{(\sqrt{2}\cos \theta )}^{2}}\] |
| \[\Rightarrow \]\[{{\cos }^{2}}\theta +{{\sin }^{2}}\theta +2\sin \theta \cos \theta =2{{\cos }^{2}}\theta \] |
| \[\Rightarrow \]\[2\sin \theta \cos \theta ={{\cos }^{2}}\theta -{{\sin }^{2}}\theta \] |
| \[\Rightarrow \]\[2\sin \theta \cos \theta =(\cos \theta -\sin \theta )(\cos \theta +\sin \theta )\] |
| \[\therefore \]\[\cos \theta -\sin \theta =\frac{2\sin \theta \cos \theta }{(\cos \theta +\sin \theta )}\] |
| \[=\frac{2\sin \theta \cos \theta }{\sqrt{2}\cos \theta }=\sqrt{2}\sin \theta \] |
You need to login to perform this action.
You will be redirected in
3 sec
