Geometric Proof that $\frac{d}{d\theta}(\tan{\theta}) = 1 + \tan^{2}{\theta}$

Question

The following proof sketch is from the preface of Tristan Needham's Visual Complex Analysis:

Why is the left side of the black triangle labeled $L \, d\theta$?

I can see that the length of this side is $L \tan{d\theta}$, so it seems that Needham is approximating $\tan{d\theta}$ with $d\theta$ as $d\theta$ approaches $0$. Why is this justified? It makes sense to me in light of the fact that the derivative of $\tan{\theta}$ at $\theta = 0$ is $1$, but this doesn't seem like the intended justification, given what he's trying to prove. Is there another, more obvious (and perhaps geometric), justification?

Answer 1

$Ld\theta$ is an arc length. You can see that the dotted line rotated $d\theta$ degrees. So, the arc length, by definition, is the radius length multiplied by the degree. In this case, $d\theta$ is too small, so we can approximate it of a straight line.

Answer 2

It was already known to ancient greek masters that $\sin \theta < \theta < \tan \theta$ (polygons inscribed and circumscribed to a circle), so I suppose that Newton,and the author here, are giving that for acquainted.