Tangled Up

The given quotient can be written as:

$\begin{aligned} Q & = \frac {\sum_{n=0}^{22} (-1)^n \binom {45}{2n+1}\tan^{2n+1}1^\circ} {\sum_{n=0}^{22} (-1)^n \binom {45}{2n}\tan^{2n}1^\circ} \\ & = \frac {\Im \left(\sum_{n=0}^{45} \binom {45}n i \tan^n 1^\circ \right)} {\Re \left(\sum_{n=0}^{45} \binom {45}n i \tan^n 1^\circ \right)} & \small \color{#3D99F6} \text{where } i = \sqrt{-1} \text{ denotes the imaginary unit, and }\Re(\cdot) \\ & = \frac {\Im \left((1+i \tan 1^\circ)^{45} \right)} {\Re \left((1+i \tan 1^\circ)^{45} \right)} & \small \color{#3D99F6} \text{and } \Im(\cdot) \text{, the real and imaginary parts respectively.} \\ & = \frac {\Im \left(\sec^{45} 1^\circ e^{i45^\circ}\right)}{\Re \left(\sec^{45} 1^\circ e^{i45^\circ}\right)} & \small \color{#3D99F6} \text{By Euler's formula: }e^{i\theta} = \cos \theta + i\sin \theta \\ & = \frac {\Im \left(e^{i45^\circ}\right)}{\Re \left(e^{i45^\circ}\right)} = \frac {\Im \left(\frac {1+i}{\sqrt 2}\right)}{\Re \left(\frac {1+i}{\sqrt 2}\right)} \\ & = \frac {\frac 1{\sqrt 2}}{\frac 1{\sqrt 2}} = \boxed 1 \end{aligned}$