Question

If $\cos \theta +\sin \theta =\sqrt{2}\cos \theta$, find $\cos \theta -\sin \theta$

• A. $\sqrt{2}\tan \theta$
• B. $\sqrt{2}\sin \theta$
• C. $\frac{\sqrt{2}}{\cos \theta +\sin \theta }$
• D. $2\sec \theta$

Answer 1

The correct option is B $\sqrt{2}\sin \theta$

Dividing by $\cos \theta$ we get,

$1+\tan \theta =\sqrt{2}\tan \theta =\sqrt{2}-1$

Hence, we can write, $\cot \theta =\frac{1}{\sqrt{2}-1}\Rightarrow \cot \theta =\frac{(\sqrt{2}+1)}{(\sqrt{2}+1)(\sqrt{2}-1)}\Rightarrow \cot \theta =\sqrt{2}+1$

So, $\frac{\cos \theta }{\sin \theta }=\sqrt{2}+1\Rightarrow \frac{\cos \theta }{\sin \theta }-1=\sqrt{2}\Rightarrow \frac{\cos \theta -\sin \theta }{\sin \theta }=\sqrt{2}\Rightarrow \cos \theta -\sin \theta =\sqrt{2}\sin \theta$

Therefore, Answer is $\sqrt{2}\sin \theta$