For the functions f(θ) = α ^2θ + β ^2θ , and g(θ) = α ^2θ + β ^2θ ,...

For the function $f(\theta) = \alpha\tan^2\theta + \beta\cot^2\theta$, applying the AM-GM inequality gives:

$\alpha\tan^2\theta + \beta\cot^2\theta \ge 2\sqrt{\alpha\tan^2\theta \cdot \beta\cot^2\theta} = 2\sqrt{\alpha\beta}$

Thus, $\min_{0 < \theta < \pi/2} f(\theta) = 2\sqrt{\alpha\beta}$.

For the function $g(\theta) = \alpha\sin^2\theta + \beta\cos^2\theta$, we can rewrite it as:

$g(\theta) = \alpha\sin^2\theta + \beta(1 - \sin^2\theta) = \beta + (\alpha - \beta)\sin^2\theta$

Since $\alpha > \beta > 0$, the maximum value of $g(\theta)$ occurs when $\sin^2\theta$ is maximum, which is $1$ (at $\theta = \pi/2$).

Thus, $\max_{0 < \theta < \pi} g(\theta) = \beta + (\alpha - \beta)(1) = \alpha$.

Given that $\min f(\theta) = \max g(\theta)$, we have:

$2\sqrt{\alpha\beta} = \alpha$

Squaring both sides (since $\alpha > 0$), we get:

$4\alpha\beta = \alpha^2 \Rightarrow \alpha = 4\beta$

For the given Geometric Progression (G.P.):

First term $a = \dfrac{\alpha}{2\beta} = \dfrac{4\beta}{2\beta} = 2$

Common ratio $r = \dfrac{2\beta}{\alpha} = \dfrac{2\beta}{4\beta} = \dfrac{1}{2}$

The sum of the first $10$ terms is:

$S_{10} = a\dfrac{1 - r^{10}}{1 - r} = 2\dfrac{1 - (1/2)^{10}}{1 - 1/2} = 4\left(1 - \dfrac{1}{1024}\right) = 4\left(\dfrac{1023}{1024}\right) = \dfrac{1023}{256}$

We are given $S_{10} = \dfrac{m}{n}$ with $\gcd(m, n) = 1$.

Since $256 = 2^8$ and $1023$ is odd, they are coprime. Therefore, $m = 1023$ and $n = 256$.

$m + n = 1023 + 256 = 1279$

Answer: $1279$