Question

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

• A. $\sqrt{2}\sin \theta$
• B. 2 $\sin \theta$
• C. $-\sqrt{2}\sin \theta$
• D. $-2\sin \theta$

Answer 1

The correct option is A $\left(\sqrt{2}\sin \theta \right)$

Given,
$\cos \theta +\sin \theta =\sqrt{2}\cos \theta$
$\sin \theta =\sqrt{2}\cos \theta -\cos \theta$=$(\sqrt{2}-1)\cos \theta$
$\Rightarrow \cos \theta =\frac{1}{(\sqrt{2}-1)}\sin \theta$
$=\sin \theta \frac{1}{\sqrt{2}-1}\times \frac{\sqrt{2}+1}{\sqrt{2}+1}$
$=\left(\sin \theta \right)\frac{\sqrt{2}+1}{{\sqrt{2}}^2-1^2}$
$=\left(\sin \theta \right)\frac{\sqrt{2}+1}{2-1}$
$\therefore \cos \theta =(\sqrt{2}+1)\sin \theta$
$\cos \theta -\sin \theta =(\sqrt{2}+1)\sin \theta -\sin \theta$
$=\sqrt{2}\sin \theta +(1-1)\sin \theta$
$\therefore \cos \theta -\sin \theta =\sqrt{2}\sin \theta$
Hence, the correct option is A $\left(\sqrt{2}\sin \theta \right)$