Let's do some calculus! (30)

$\begin{aligned} L & = \lim_{x \to 0} \frac {\ln \left(\color{#3D99F6}{\cot \left( \frac \pi 4 - k_1x \right)} \right)}{\tan \left(k_2 x\right)} & \small \color{#3D99F6}{\text{Note that }\cot x = \tan \left(\frac \pi 2-x\right)} \\ & = \lim_{x \to 0} \frac {\ln \left(\color{#3D99F6}{\tan \left( \frac \pi 4 + k_1x \right)}\right)}{\tan \left(k_2 x\right)} & \small \color{#3D99F6}{\text{This is a 0/0 case and L'Hôpital's rule applies.}} \\ & = \lim_{x \to 0} \frac {\frac {k_1\sec^2 \left( \frac \pi 4 + k_1x \right)}{\tan \left( \frac \pi 4 + k_1x \right)}}{k_2\sec^2 \left(k_2 x\right)} & \small \color{#3D99F6}{\text{Differentiate up and down w.r.t. }x} \\ & = \frac {\frac {2k_1}1}{k_2} = \frac {2k_1}{k_2} \end{aligned}$

$\begin{aligned} \implies \frac {2k_1}{k_2} & = 1 \\ \frac {k_2}{k_1} & = \boxed{2} \end{aligned}$