Quantitative Aptitude Ques 916
Question: If $ \cos \theta +\sin \theta =\sqrt{2}\cos \theta . $ Then, the value of $ \cos \theta -\sin \theta $ is
Options:
A) $ \sqrt{3}\sin \theta $
B) $ \sqrt{2}\cos \theta $
C) $ \sqrt{2}\sin \theta $
D) $ \sqrt{3}\cos \theta $
Show Answer
Answer:
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 $
BETA
