The minimum value of 5 tan^2 alpha + frac{9}{ | vedclass

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(B) Given expression: $5 	an^2 \alpha + \frac{9}{	an^2 \alpha} + 4 \sec^2 \alpha$ Using the identity $\sec^2 \alpha = 1 + 	an^2 \alpha$,we get: $=

Vedclass · Accepted Answer

$22$

Vedclass · Answer

(B) Given expression: $5 	an^2 \alpha + \frac{9}{	an^2 \alpha} + 4 \sec^2 \alpha$ Using the identity $\sec^2 \alpha = 1 + 	an^2 \alpha$,we get: $= 5 	an^2 \alpha + 9 \cot^2 \alpha + 4(1 + 	an^2 \alpha)$ $= 9 	an^2 \alpha + 9 \cot^2 \alpha + 4$ Since $AM \geq GM$ for positive real numbers,we have: $\frac{9 	an^2 \alpha + 9 \cot^2 \alpha}{2} \geq \sqrt{9 	an^2 \alpha \cdot 9 \cot^2 \alpha}$ $9 	an^2 \alpha + 9 \cot^2 \alpha \geq 2 \cdot \sqrt{81 \cdot 1}$ $9 	an^2 \alpha + 9 \cot^2 \alpha \geq 18$ Adding $4$ to both sides: $9 	an^2 \alpha + 9 \cot^2 \alpha + 4 \geq 18 + 4 = 22$ Therefore,the minimum value is $22$.

The minimum value of $5 \tan^2 \alpha + \frac{9}{\tan^2 \alpha} + 4 \sec^2 \alpha$ is:

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