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

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Question

(A) Let $f(	heta) = 8 \sec^2 	heta + 2 \cos^2 	heta$.Using the Arithmetic Mean-Geometric Mean ($AM$-$GM$) inequality,for positive real numbers $a$

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(A) Let $f(	heta) = 8 \sec^2 	heta + 2 \cos^2 	heta$.Using the Arithmetic Mean-Geometric Mean ($AM$-$GM$) inequality,for positive real numbers $a$ and $b$,the minimum value of $a + b$ is $2\sqrt{ab}$.Here,$a = 8 \sec^2 	heta$ and $b = 2 \cos^2 	heta$.$f(	heta) \ge 2 \sqrt{(8 \sec^2 	heta) \cdot (2 \cos^2 	heta)}$.Since $\sec^2 	heta \cdot \cos^2 	heta = 1$,we have:$f(	heta) \ge 2 \sqrt{16 \cdot 1} = 2 \cdot 4 = 8$.However,we must check if this minimum is attainable. The equality in $AM$-$GM$ holds when $a = b$,i.e.,$8 \sec^2 	heta = 2 \cos^2 	heta$.$4 \sec^2 	heta = \cos^2 	heta \implies 4 = \cos^4 	heta \implies \cos^2 	heta = 2$,which is impossible as $\cos^2 	heta \le 1$.Thus,we evaluate the function $f(	heta) = 8(1 + 	an^2 	heta) + 2 \cos^2 	heta$. Since $	an^2 	heta \ge 0$,the minimum occurs at $	an 	heta = 0$,i.e.,$	heta = 0^\circ$.At $	heta = 0^\circ$,$f(0) = 8 \sec^2(0) + 2 \cos^2(0) = 8(1) + 2(1) = 10$.

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

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