That is a long fraction!

$\begin{aligned} \frac 2{\sin t - \frac 2{\sin t - \frac 2{\sin t - \cdots}}} & = \pi \\ \frac 2{\sin t - \pi} & = \pi \\ \implies \sin t & = \frac 2\pi + \pi \\ \frac {e^{it}-e^{-it}}{2i} & = \frac 2\pi + \pi \\ e^{2it} - 2i \left(\frac 2\pi + \pi \right) e^{it} - 1 & = 0 \end{aligned}$

$\begin{aligned} \ \implies e^{it} & = \frac 12 \left(2i\left(\frac 2\pi + \pi \right) + \sqrt{-4\left(\frac 2\pi + \pi \right)^2+4}\right) \\ & = \left(\frac 2\pi + \pi + \sqrt{\left(\frac 2\pi + \pi \right)^2-1}\right) i \\ & = \left(\frac 2\pi + \pi + \sqrt{\left(\frac 2\pi + \pi \right)^2-1}\right) e^{\left(2\blue n\pi + \frac \pi 2\right)i} & \small \blue{\text{where }n \text{ is an integer.}} \\ it & = \ln \left(\frac 2\pi + \pi + \sqrt{\left(\frac 2\pi + \pi \right)^2-1}\right) + \left(2n\pi + \frac \pi 2\right)i \\ t & = 2n \pi + \frac \pi 2 - i \ln \left(\frac 2\pi + \pi + \sqrt{\left(\frac 2\pi + \pi \right)^2-1}\right) \end{aligned}$

Therefore, $\dfrac {a+b}b = \dfrac {-1+2}2 = \boxed{0.5}$ .

I guessed and somehow got the correct answer with what I believe to be my incorrect solution. Can someone inform me as to what I am missing?

The given fraction can be represented as $\pi = \frac{2}{\sin(t) - \pi}$ and, solving for $\sin(t)$ , we get $\sin\left(t\right)=\frac{2}{\pi}+\pi$ . We can also use this fact to find $\cos\left(t\right)=\sqrt{1-\left(\frac{2}{\pi}+\pi\right)^{2}}$ . Using Euler's Formula, $e^{it}=\cos\left(t\right)+i\sin\left(t\right)=\sqrt{1-\left(\frac{2}{\pi}+\pi\right)^{2}}+i\left(\frac{2}{\pi}+\pi\right) = i\left(\sqrt{\left(\frac{2}{\pi}+\pi\right)^{2}-1}+\left(\frac{2}{\pi}+\pi\right)\right)$ . Solving for t, we get $it = \ln i\left(i\sqrt{\left(\frac{2}{\pi}+\pi\right)^{2}-1}+\frac{2}{\pi}+\pi\right)$ or $t = -i\ln i\left(i\sqrt{\left(\frac{2}{\pi}+\pi\right)^{2}-1}+\frac{2}{\pi}+\pi\right)$ . From this point, I guessed that $a = -1$ and $b = 2$ , with an answer of $\boxed{\frac{1}{2}}$ .

Note that Wolfram Alpha gives the same result as above, alongside equivalent solution $t = \pi/2 - i\ln(\frac{2}{\pi} + \pi + \sqrt{-1 + (\frac{2}{\pi} + \pi)^2})$ .