The minimum value of
Question:

The minimum value of 9 tan2θ + 4 cot2θ is ____________.

Solution:

9 tan2θ + 4 cot2θ

Since Arithmetic mass ≥ Geometric mean for 2 tans.

i. e $\frac{a+b}{2} \geq \sqrt{a b}$

Let $a=9 \tan ^{2} \theta$

$b=4 \cot ^{2} \theta$

$\Rightarrow \frac{9 \tan ^{2} \theta+4 \cot ^{2} \theta}{2} \geq \sqrt{9 \tan ^{2} \theta \times 4 \cot ^{2} \theta}$

$=\sqrt{9 \times 4}$

$=3 \times 2$

i. e $9 \tan ^{2} \theta+4 \cot ^{2} \theta \geq 2 \times 6=12$

i.e minimum value of $9 \tan ^{2} \theta+4 \cot ^{2} \theta$ is 12 .