The minimum value of 9tan^2theta + 4cot^2thet | 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 that $\frac{a+b}{2} \ge \sqrt

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 that $\frac{a+b}{2} \ge \sqrt{ab}$.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	heta \cdot \cot	heta = 1$,we have $\sqrt{9 \cdot 4 \cdot 1} = \sqrt{36} = 6$.Therefore,$\frac{9	an^2	heta + 4\cot^2	heta}{2} \ge 6$.Multiplying both sides by $2$,we get $9	an^2	heta + 4\cot^2	heta \ge 12$.Thus,the minimum value is $12$.

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

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