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

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(B) Given: $\sin 	heta + \cos 	heta = \sqrt{2} \cos 	heta$ .... $(1)$Let $x = \cos 	heta - \sin 	heta$ .... $(2)$Squaring both equations:$(\sin \

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$\sqrt{2} \sin 	heta$

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(B) Given: $\sin 	heta + \cos 	heta = \sqrt{2} \cos 	heta$ .... $(1)$Let $x = \cos 	heta - \sin 	heta$ .... $(2)$Squaring both equations:$(\sin 	heta + \cos 	heta)^2 = (\sqrt{2} \cos 	heta)^2 \Rightarrow \sin^2 	heta + \cos^2 	heta + 2 \sin 	heta \cos 	heta = 2 \cos^2 	heta$$1 + 2 \sin 	heta \cos 	heta = 2 \cos^2 	heta$ .... $(3)$$x^2 = (\cos 	heta - \sin 	heta)^2 = \cos^2 	heta + \sin^2 	heta - 2 \sin 	heta \cos 	heta$$x^2 = 1 - 2 \sin 	heta \cos 	heta$ .... $(4)$Adding $(3)$ and $(4)$:$1 + 2 \sin 	heta \cos 	heta + x^2 = 2 \cos^2 	heta + 1 - 2 \sin 	heta \cos 	heta$$1 + x^2 = 2 \cos^2 	heta + 1$$x^2 = 2 \cos^2 	heta$Wait,let's re-evaluate: From $(1)$,$\sin 	heta = (\sqrt{2} - 1) \cos 	heta$.Then $\cos 	heta - \sin 	heta = \cos 	heta - (\sqrt{2} - 1) \cos 	heta = \cos 	heta (1 - \sqrt{2} + 1) = \cos 	heta (2 - \sqrt{2}) = \sqrt{2} \cos 	heta (\sqrt{2} - 1)$.Alternatively,using the identity $(\sin 	heta + \cos 	heta)^2 + (\cos 	heta - \sin 	heta)^2 = 2(\sin^2 	heta + \cos^2 	heta) = 2$.$(\sqrt{2} \cos 	heta)^2 + x^2 = 2$$2 \cos^2 	heta + x^2 = 2$$x^2 = 2(1 - \cos^2 	heta) = 2 \sin^2 	heta$$x = \sqrt{2} \sin 	heta$.

If $\sin \theta + \cos \theta = \sqrt{2} \cos \theta$,then the value of $(\cos \theta - \sin \theta)$ is

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