How is $ds/d\theta =R_0/\cos^2\Psi$ derived?

Question

I have a textbook on general relativity that states the following:

The general straight line is:

$r\cos( \theta - a ) = R_0$

where $a$ and $R_0$ are arbitrary constants. This equation, and $ds^2 = dr^2 + r^2d\theta^2$, give $ds/d\theta =R_0/\cos^2\Psi$ where $\Psi=\theta-a$.

However, I can't figure out the steps to derive that result.

Answer 1

We have $$r=\frac{R_0}{\cos(\theta-a)} \implies dr = \frac{R_0}{\cos^2(\theta-a)}\cdot\sin(\theta-a)d\theta \implies \frac{dr^2}{d\theta^2}=\frac{R_0^2\sin^2(\theta-a)}{\cos^4(\theta-a)}$$

Now, $$\frac{ds^2}{d\theta^2} = \frac{dr^2}{d\theta^2}+r^2=\frac{R_0^2\sin^2(\theta-a)}{\cos^4(\theta-a)}+\frac{R_0^2}{\cos^2(\theta-a)}=\frac{R_0^2}{\cos^4\psi}\left(\sin^2\psi+\cos^2\psi\right)=\frac{R_0^2}{\cos^4\psi}$$

$$\frac{ds}{d\theta}=\frac{R_0}{\cos^2\psi}$$