The minimum value of $a \tan ^2 x+b \cot ^2 x$ equals the maximum value of $a \sin ^2 \theta+b \cos ^2 \theta$, where $a>b>0$, when 

$a=4 b$

Let $y=a \tan ^2 x+b \cot ^2 x $ and $z=a \sin ^2 \theta+b \cos ^2 \theta$. Then,

$y=(\sqrt{a} \tan x-\sqrt{b} \cot x)^2+2 \sqrt{a b} \geq 2 \sqrt{a b}$

$\Rightarrow y_{\min }=2 \sqrt{a b}$

and, $z=a \sin ^2 \theta+b \cos ^2 \theta$

$\frac{d z}{d \theta}=(a-b) \sin 2 \theta$ and $\frac{d^2 z}{d \theta^2}=2(a-b) \cos 2 \theta$

For z to be maximum/minimum, we must have

$\frac{d z}{d \theta}=0 \Rightarrow \sin 2 \theta=0 \Rightarrow 2 \theta=0, \pi, 2 \pi \Rightarrow \theta=0, \pi / 2, \pi$

Clearly, $\frac{d^2 z}{d \theta^2}<0$ for $\theta=\frac{\pi}{2}$. Hence, z is maximum for $\theta=\frac{\pi}{2}$.

∴  $z_{\max }=a \sin ^2 \frac{\pi}{2}+b \cos ^2 \frac{\pi}{2}=a$

Now,

$y_{\min }=z_{\max } \Rightarrow 2 \sqrt{a b}=a \Rightarrow 2 \sqrt{b}=\sqrt{a} \Rightarrow a=4 b$