If $\cos \theta+\sin \theta=\sqrt{2} \cos \theta$, show that $\cos \theta-\sin \theta=\sqrt{2} \sin \theta .$
Given that if $\cos \theta+\sin \theta=\sqrt{2} \cos \theta$, then we have to prove that $\cos \theta-\sin \theta=\sqrt{2} \sin \theta$
We have,
$\cos \theta+\sin \theta=\sqrt{2} \cos \theta$
$\Rightarrow \quad \sin \theta=\sqrt{2} \cos \theta-\cos \theta$
$\Rightarrow \quad \sin \theta=(\sqrt{2}-1) \cos \theta$
$\Rightarrow \quad \sin \theta=\frac{(\sqrt{2}-1)(\sqrt{2}+1)}{\sqrt{2}} \cos \theta$
$\Rightarrow \quad \sin \theta=\frac{\cos \theta}{\sqrt{2}+1}$
$\Rightarrow \quad \cos \theta=\sqrt{2} \sin \theta+\sin \theta$
$\Rightarrow \cos \theta-\sin \theta=\sqrt{2} \sin \theta$
Hence proved.
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