Question

\[ 9\tan^2\theta+4\cot^2\theta \]

\[ \text{Find the minimum value} \]

Solution

Let

\[ \tan^2\theta=x \]

Then

\[ \cot^2\theta=\frac1x \]

So,

\[ 9\tan^2\theta+4\cot^2\theta = 9x+\frac4x \]

Using AM ≥ GM,

\[ 9x+\frac4x \ge 2\sqrt{9x\cdot\frac4x} \]

\[ = 2\sqrt{36} =12 \]

Hence minimum value is

\[ 12 \]

Answer

\[ \boxed{12} \]