Physics based trigonometry question

Question

I am working on solving the problem stated in this image:

I understand almost everything about this problem. I solved for the magnetic field along the axis of a circular loop, and now I need to integrate from Ө1 to Ө2.

The problem I am encountering is that I am not sure how to put dz in terms of dӨ. Here is how the example does it:

Can someone explain these steps to me? I pretty much understand every step here except for this one:

$$ \frac{d\theta}{\cos^2\theta} = -\frac a{z^2}\,dz $$

Thanks for your help!

Answer 1

Given:

$$ \tan \theta = \frac az$$

Now perform differentiation on both sides against some dummy variable $t$:

\begin{align} \frac{d}{dt}\tan\theta &= \frac{d\theta}{dt} \sec^2\theta \\ \frac{d}{dt}\frac az &= -\frac{dz}{dt} \frac{a}{z^2}. \end{align}

"Eliminate" the $dt$ and use $\sec\theta=\frac1{\cos\theta}$.

$$\frac{d\theta}{\cos^2\theta} = -\frac{a}{z^2}dz.$$

Answer 2

$$ \tan\theta = \frac a z \tag 1 $$ Consequently $$ \frac 1 {\cos^2\theta} = \frac d {d\theta} \tan \theta = \underbrace{\frac d {d\theta} \frac a z = \frac{-a}{z^2}\,\frac{dz}{d\theta}}_{\text{chain rule}} $$ $$ \frac{d\theta}{\cos^2\theta} = \frac{-a}{z^2}\,dz $$

Answer 3

Differentiate $$ z \tan \theta = a \tag{1} $$ $$ dz \tan \theta + z \sec ^2\theta\; d\theta = 0 \tag{2}$$ $$ \dfrac{d \theta }{dz}=\dfrac{-\tan\theta \cos^2 \theta}{z}= \dfrac{(-a/z) \cos^2 \theta}{z}=\dfrac{-a \cos^2 \theta}{z^2} \tag{3}.$$ and re-arrange.