The value of cos ^{2} theta+sec ^{2} theta is | vedclass

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(D) We can write the expression as: $\cos ^{2} 	heta+\sec ^{2} 	heta = (\cos 	heta - \sec 	heta)^{2} + 2 \cos 	heta \sec 	heta$.Since $\cos 	he

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Greater than or equal to $2$

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(D) We can write the expression as: $\cos ^{2} 	heta+\sec ^{2} 	heta = (\cos 	heta - \sec 	heta)^{2} + 2 \cos 	heta \sec 	heta$.Since $\cos 	heta \sec 	heta = 1$,the expression becomes: $(\cos 	heta - \sec 	heta)^{2} + 2$.Because the square of any real number is always non-negative,$(\cos 	heta - \sec 	heta)^{2} \ge 0$.Therefore,$(\cos 	heta - \sec 	heta)^{2} + 2 \ge 2$.Thus,the value of $\cos ^{2} 	heta+\sec ^{2} 	heta$ is always greater than or equal to $2$.

The value of $\cos ^{2} \theta+\sec ^{2} \theta$ is always

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