If cos theta + sin theta = sqrt{2} cos theta, | vedclass

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(A) Given the equation: $\cos 	heta + \sin 	heta = \sqrt{2} \cos 	heta$Rearranging the terms to isolate $\sin 	heta$:$\sin 	heta = \sqrt{2} \cos

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$\sqrt{2} \sin 	heta$

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(A) Given the equation: $\cos 	heta + \sin 	heta = \sqrt{2} \cos 	heta$Rearranging the terms to isolate $\sin 	heta$:$\sin 	heta = \sqrt{2} \cos 	heta - \cos 	heta$$\sin 	heta = (\sqrt{2} - 1) \cos 	heta$Now,solve for $\cos 	heta$:$\cos 	heta = \frac{\sin 	heta}{\sqrt{2} - 1}$Rationalizing the denominator by multiplying the numerator and denominator by $(\sqrt{2} + 1)$:$\cos 	heta = \frac{(\sqrt{2} + 1) \sin 	heta}{(\sqrt{2} - 1)(\sqrt{2} + 1)}$$\cos 	heta = \frac{(\sqrt{2} + 1) \sin 	heta}{2 - 1}$$\cos 	heta = (\sqrt{2} + 1) \sin 	heta$$\cos 	heta = \sqrt{2} \sin 	heta + \sin 	heta$Finally,calculate $\cos 	heta - \sin 	heta$:$\cos 	heta - \sin 	heta = \sqrt{2} \sin 	heta$

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

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