If 7 sin^{2} theta + 3 cos^{2} theta = 4,then | vedclass

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(D) Given equation: $7 \sin^{2} 	heta + 3 \cos^{2} 	heta = 4$.Using the identity $\cos^{2} 	heta = 1 - \sin^{2} 	heta$,we substitute:$7 \sin^{2} \

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$1/\sqrt{3}$

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(D) Given equation: $7 \sin^{2} 	heta + 3 \cos^{2} 	heta = 4$.Using the identity $\cos^{2} 	heta = 1 - \sin^{2} 	heta$,we substitute:$7 \sin^{2} 	heta + 3(1 - \sin^{2} 	heta) = 4$$7 \sin^{2} 	heta + 3 - 3 \sin^{2} 	heta = 4$$4 \sin^{2} 	heta = 4 - 3$$4 \sin^{2} 	heta = 1$$\sin^{2} 	heta = 1/4$Since $	heta$ is acute,$\sin 	heta = 1/2$.Thus,$	heta = 30^{\circ}$.Therefore,$	an 	heta = 	an 30^{circ} = 1/\sqrt{3}$.

If $7 \sin^{2} \theta + 3 \cos^{2} \theta = 4$,then the value of $\tan \theta$ is (where $\theta$ is acute).

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