The minimum value of 9 tan^2 theta + 4 cot^2 | vedclass

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(D) We use the Arithmetic Mean-Geometric Mean $(AM \ge GM)$ inequality for two positive numbers $a$ and $b$,which states $\frac{a+b}{2} \ge \sqrt{ab}$

Vedclass · Accepted Answer

$12$

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(D) We use the Arithmetic Mean-Geometric Mean $(AM \ge GM)$ inequality for two positive numbers $a$ and $b$,which states $\frac{a+b}{2} \ge \sqrt{ab}$.Let $a = 9 	an^2 	heta$ and $b = 4 \cot^2 	heta$.Then,$\frac{9 	an^2 	heta + 4 \cot^2 	heta}{2} \ge \sqrt{9 	an^2 	heta \cdot 4 \cot^2 	heta}$.Since $	an^2 	heta \cdot \cot^2 	heta = 1$,we have $\frac{9 	an^2 	heta + 4 \cot^2 	heta}{2} \ge \sqrt{36}$.$\frac{9 	an^2 	heta + 4 \cot^2 	heta}{2} \ge 6$.$9 	an^2 	heta + 4 \cot^2 	heta \ge 12$.Therefore,the minimum value is $12$.

The minimum value of $9 \tan^2 \theta + 4 \cot^2 \theta$ is

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