A Very Tanned Log

$\cos x=\sin(90-x)$

So the logs $\tan(x)$ and $\tan{90-x}$ have the same base.

Using the log property $\log_ca+\log_cb=\log_cab$

We get $\displaystyle \sum_1^{89} \log_{\cos{x}} \left({\tan{x}*\tan{(90-x)}}\right)$

$\log_{\sin45}\left(\tan 45\right)+\displaystyle \sum_1^{44} \log_{\cos{x}}\left({\dfrac{\sin x}{\cos x}*\dfrac{\sin{(90-x)}}{\cos{(90-x)}}}\right)$

$0+\displaystyle \sum_1^{44} \log_{\cos{x}} \left({\dfrac{\sin x}{\cos x}*\dfrac{\cos{x}}{\sin{x}}}\right)$

$\displaystyle \sum_1^{44} \log_{\cos{x}}1$

Regardless of what cos x is, each log will be 0. Thus our total sum is 0.

Let the sum of the expression be $S$ , then we have:

$\begin{aligned} S & = \sum_{k=1}^{44} \log_{\cos k^\circ} ( \tan k^\circ) + \sum_{k=45}^{89} \log_{\sin k^\circ} (\tan k^\circ ) \\ & = \sum_{k=1}^{44} \frac{\log ( \tan k^\circ)}{\log (\cos k^\circ)} + \sum_{k=45}^{89} \frac{\log ( \tan k^\circ)}{\log (\sin k^\circ)} \\ & = \sum_{k=1}^{44} \frac{\log (\sin k^\circ) - \log (\cos k^\circ)}{\log (\cos k^\circ)} + \sum_{k=45}^{89} \frac{\log (\sin k^\circ) - \log (\cos k^\circ)}{\log (\sin k^\circ)} \\ & = \sum_{k=1}^{44} \frac{\log (\sin k^\circ) - \log (\cos k^\circ)}{\log (\cos k^\circ)} + \sum_{k=\color{#3D99F6}{45}}^{\color{#3D99F6}{89}} \frac{\log (\color{#3D99F6}{\cos (90- k)^\circ}) - \log (\color{#3D99F6}{\sin (90 - k)^\circ)}}{\log (\color{#3D99F6}{\cos (90- k)^\circ})} \\ & = \sum_{k=1}^{44} \frac{\log (\sin k^\circ) - \log (\cos k^\circ)}{\log (\cos k^\circ)} + \sum_{k=\color{#3D99F6}{1}}^{\color{#3D99F6}{44}} \frac{\log (\color{#3D99F6}{\cos k^\circ}) - \log (\color{#3D99F6}{\sin k^\circ)}}{\log (\color{#3D99F6}{\cos k^\circ})} \\ & = \boxed{0} \end{aligned}$