Answer
$\cot\theta=\frac{L_{0}-L_{1}}{s} $
Work Step by Step
We use teh identity:
$$\cot\theta=\frac{1}{\tan\theta}$$
So using the definition of tangent, in $\triangle BDC$ we have:
$$\tan\theta=\dfrac{BC}{BD}=\dfrac{s}{L_{0}-L_{1}}$$
so:
$$\cot\theta=\frac{1}{\tan\theta}=\frac{1}{\frac{s}{L_{0}-L_{1}}}=\frac{L_{0}-L_{1}}{s} $$
