Texas

I used a quite different approach. I realized that the function have when I figured that $f(45) > f(60) < f(75)$ in degree, where $f(a)$ is the given function.

Let $f(a) = sec(a) + 8cosec(a)$ . Then $f'(a) = \frac{sin^{3}a - 8cos^{3}a}{sin^{2}(a)cos^{2}(a)}$ .

Since we're looking for the minimum value of $f(a)$ , then we just need to find $x \in (45, 75)$ where $f'(x) = 0$ .

It if obvious that $f'(x) = \frac{(sin(x) - 2 cos(x))(sin^{2}x + 2sin(x)cos(x) + 4cos^{2}x)}{sin^{2}(x)cos^{2}(x)}$ .

Also,

$sin^{2}a + 4cos^{2}a + sin2a > 0 \forall a \in (0, 90)$ ,

and

$sin^{2}(a)cos^{2}(a) \geq 0 \forall a \in R$ .

Therefore,

$sin(a) - 2cos(a) = 0 <=> tan(a) = 2$ .

Thus $sin(a) = \frac{2}{\sqrt{5}}, cos(a) = \frac{1}{\sqrt{5}}$ .

Thus, $k = \sqrt{5} + 4\sqrt{5} = 5\sqrt{5} = 5^{\frac{3}{2}} = a^{b} <=> a + b = \frac{13}{2}$ .

The title is supposed to be Texas Hold 'Em, in which Hold 'Em is supposed to be a pun for Holder's. Because a pun or two won't hurt occasionally. :)

Since $0 < \theta < \frac{\pi}{2}$ , this implies that both $\sec \theta, \csc \theta$ are positive. Moreover, $\frac{1}{\sec^2 \theta}+\frac{1}{\csc^2 \theta} = \cos^2 \theta + \sin^2 \theta = 1$ . Since $\sec \theta + 8 \csc \theta = k$ will have exactly one solution for $\theta$ when $k$ is the minimu value of $\sec \theta + 8 \csc \theta$ , we just need to find the minimum value of $\sec \theta + 8 \csc \theta$ .

By Holder's Inequality,

$\large (\sec \theta + 8 \csc \theta)^{\frac{2}{3}}(\frac{1}{\sec^2 \theta}+\frac{1}{\csc^2 \theta})^{\frac{1}{3}} \geq (\sec \theta)^{\frac{2}{3}} (\frac{1}{\sec^2 \theta})^{\frac{1}{3}} +(8 \csc \theta)^{\frac{2}{3}} (\frac{1}{\csc^2 \theta})^{\frac{1}{3}} = 1+ 8^\frac{2}{3} = 5$

But since $\large \frac{1}{\sec^2 \theta} +\frac{1}{\csc^2 \theta} = 1$ , then $\large (\sec \theta + 8 \csc \theta)^{\frac{2}{3}} \geq 5$ . This implies that $\large \sec \theta + 8 \csc \theta \geq 5^\frac{3}{2}$ .

So, $a = 5, b = \frac{3}{2}$ . Therefore our answer is $a+b = \boxed{6.5}$ .