If $cos^{2} θ + se c^{2} θ = p$ , then

p < 1

p = 1

1 < p < 2

p \geq 2

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Elijah Discussed

cos^{2} θ + se c^{2} θ = p

, then

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Text solutionVerified

p = (cos^{2} θ + se c^{2} θ) = (cos θ - sec θ)^{2} + 2 cos θ sec θ = (cos θ - sec θ)^{2} + 2 \geq 2

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