If cos theta - 4 sin theta = 1,then sin theta | vedclass

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(D) Let $x = \cos 	heta - 4 \sin 	heta = 1$ and $y = \sin 	heta + 4 \cos 	heta$. Squaring both equations and adding them: $x^2 + y^2 = (\cos 	het

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$\pm 4$

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(D) Let $x = \cos 	heta - 4 \sin 	heta = 1$ and $y = \sin 	heta + 4 \cos 	heta$. Squaring both equations and adding them: $x^2 + y^2 = (\cos 	heta - 4 \sin 	heta)^2 + (\sin 	heta + 4 \cos 	heta)^2$ $x^2 + y^2 = (\cos^2 	heta + 16 \sin^2 	heta - 8 \sin 	heta \cos 	heta) + (\sin^2 	heta + 16 \cos^2 	heta + 8 \sin 	heta \cos 	heta)$ $x^2 + y^2 = \cos^2 	heta + \sin^2 	heta + 16(\sin^2 	heta + \cos^2 	heta)$ $x^2 + y^2 = 1 + 16(1) = 17$ Since $x = 1$,we have $1^2 + y^2 = 17$ $y^2 = 16$ $y = \pm 4$ Therefore,$\sin 	heta + 4 \cos 	heta = \pm 4$.

If $\cos \theta - 4 \sin \theta = 1$,then $\sin \theta + 4 \cos \theta$ is equal to

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