If $(\cos \theta-\sin \theta)=\sqrt{2} \sin \theta$ then prove that $(\cos \theta+\sin \theta)=\sqrt{2} \cos \theta$.
$\cos \theta-\sin \theta=\sqrt{2} \sin \theta$
Squaring on both sides, we get
$(\cos \theta-\sin \theta)^{2}=(\sqrt{2} \sin \theta)^{2}$
$\Rightarrow \cos ^{2} \theta+\sin ^{2} \theta-2 \sin \theta \cos \theta=2 \sin ^{2} \theta$
$\Rightarrow \cos ^{2} \theta-\sin ^{2} \theta=2 \sin \theta \cos \theta$
$\Rightarrow(\cos \theta-\sin \theta)(\cos \theta+\sin \theta)=2 \sin \theta \cos \theta \quad\left[a^{2}-b^{2}=(a-b)(a+b)\right]$
$\Rightarrow \sqrt{2} \sin \theta(\cos \theta+\sin \theta)=2 \sin \theta \cos \theta \quad(\cos \theta-\sin \theta=\sqrt{2} \sin \theta)$
$\Rightarrow \cos \theta+\sin \theta=\frac{2 \sin \theta \cos \theta}{\sqrt{2} \sin \theta}$
$\Rightarrow \cos \theta+\sin \theta=\sqrt{2} \cos \theta$
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