If 3cos theta + 4sin theta = 5,then the value | vedclass

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(C) Let $3\cos 	heta + 4\sin 	heta = 5$ be equation $(1)$ and $3\sin 	heta - 4\cos 	heta = x$ be equation $(2)$.Squaring both equations and adding

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$0$

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(C) Let $3\cos 	heta + 4\sin 	heta = 5$ be equation $(1)$ and $3\sin 	heta - 4\cos 	heta = x$ be equation $(2)$.Squaring both equations and adding them:$(3\cos 	heta + 4\sin 	heta)^2 + (3\sin 	heta - 4\cos 	heta)^2 = 5^2 + x^2$$(9\cos^2 	heta + 16\sin^2 	heta + 24\sin 	heta \cos 	heta) + (9\sin^2 	heta + 16\cos^2 	heta - 24\sin 	heta \cos 	heta) = 25 + x^2$$9(\cos^2 	heta + \sin^2 	heta) + 16(\sin^2 	heta + \cos^2 	heta) = 25 + x^2$Since $\sin^2 	heta + \cos^2 	heta = 1$,we get:$9(1) + 16(1) = 25 + x^2$$25 = 25 + x^2$$x^2 = 0$Therefore,$x = 0$.

If $3\cos \theta + 4\sin \theta = 5$,then the value of $3\sin \theta - 4\cos \theta$ is:

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