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 .