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

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(C) Let $3 \cos 	heta + 4 \sin 	heta = 5$  $(i)$ Let $3 \sin 	heta - 4 \cos 	heta = x$  $(ii)$ Squaring both equations and adding them: $(3 \cos \

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(C) Let $3 \cos 	heta + 4 \sin 	heta = 5$  $(i)$ Let $3 \sin 	heta - 4 \cos 	heta = x$  $(ii)$ 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 have: $9(1) + 16(1) = 25 + x^2$ $25 = 25 + x^2$ $x^2 = 0$ 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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