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

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(C) Given equation: $7 \sin ^{2} 	heta+3 \cos ^{2} 	heta=4$Divide the entire equation by $\cos ^{2} 	heta$:$7 \frac{\sin ^{2} 	heta}{\cos ^{2} 	h

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$\pm \frac{1}{\sqrt{3}}$

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(C) Given equation: $7 \sin ^{2} 	heta+3 \cos ^{2} 	heta=4$Divide the entire equation by $\cos ^{2} 	heta$:$7 \frac{\sin ^{2} 	heta}{\cos ^{2} 	heta} + 3 \frac{\cos ^{2} 	heta}{\cos ^{2} 	heta} = \frac{4}{\cos ^{2} 	heta}$Using the identities $	an 	heta = \frac{\sin 	heta}{\cos 	heta}$ and $\sec 	heta = \frac{1}{\cos 	heta}$:$7 	an ^{2} 	heta + 3 = 4 \sec ^{2} 	heta$Substitute $\sec ^{2} 	heta = 1 + 	an ^{2} 	heta$:$7 	an ^{2} 	heta + 3 = 4(1 + 	an ^{2} 	heta)$Expand the right side:$7 	an ^{2} 	heta + 3 = 4 + 4 	an ^{2} 	heta$Rearrange the terms to solve for $	an ^{2} 	heta$:$7 	an ^{2} 	heta - 4 	an ^{2} 	heta = 4 - 3$$3 	an ^{2} 	heta = 1$$	an ^{2} 	heta = \frac{1}{3}$Taking the square root on both sides:$	an 	heta = \pm \frac{1}{\sqrt{3}}$

If $7 \sin ^{2} \theta+3 \cos ^{2} \theta=4,$ then $\tan \theta=$

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