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7^2 + 3^2 = 4, then

Question

$7 sin^{2} θ + 3 cos^{2} θ = 4$ , then

tan θ = \frac{1}{2}

tan θ = \frac{1}{2}

tan θ = \frac{1}{3}

tan θ = \frac{1}{3}

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Mila Discussed

7 sin^{2} θ + 3 cos^{2} θ = 4

, then

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Text solutionVerified

7 s i n^{2} θ + 3 co s^{2} θ = 4

4 s i n^{2} θ + 3 (co s^{2} θ + s i n^{2} θ) = 4

4 s i n^{2} θ + 3 = 4

4 s i n^{2} θ = 1

s i n^{2} θ = \frac{1}{4}

s in θ = \pm \frac{1}{2}

Considering

s in θ = \frac{1}{2}

θ = 3 0^{0}

Hence

t an θ = t an 3 0^{0} = \frac{1}{3}

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