The minimum value of (8 sec^2 theta + 2 cos^2 | vedclass

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Question

(C) Let $f(	heta) = 8 \sec^2 	heta + 2 \cos^2 	heta$. Using the Arithmetic Mean-Geometric Mean $(AM \ge GM)$ inequality for positive terms: $\frac{

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(C) Let $f(	heta) = 8 \sec^2 	heta + 2 \cos^2 	heta$. Using the Arithmetic Mean-Geometric Mean $(AM \ge GM)$ inequality for positive terms: $\frac{8 \sec^2 	heta + 2 \cos^2 	heta}{2} \ge \sqrt{8 \sec^2 	heta 	imes 2 \cos^2 	heta}$ $\frac{8 \sec^2 	heta + 2 \cos^2 	heta}{2} \ge \sqrt{16 \sec^2 	heta \cos^2 	heta}$ Since $\sec^2 	heta \cos^2 	heta = 1$,we have: $\frac{8 \sec^2 	heta + 2 \cos^2 	heta}{2} \ge \sqrt{16}$ $8 \sec^2 	heta + 2 \cos^2 	heta \ge 2 	imes 4$ $8 \sec^2 	heta + 2 \cos^2 	heta \ge 8$. Thus,the minimum value is $8$.

The minimum value of $(8 \sec^2 \theta + 2 \cos^2 \theta)$ is equal to :-

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