I've drawn the below diagram.
$$ dx= a \cdot \sec^{2}\left( \phi_{} \right) d \phi_{} ~~ \leftarrow~~ \text{?}$$
$$ x= a \tan \left( \phi_{} \right) ~~ \leftarrow~~ \text{ok} $$
$$ \phi_{} = \frac{\pi}{2} -\theta_{} ~~ \leftarrow~~ \text{ok} $$
How do I deduce it?
My tries are as below.
$$ \frac{ dx }{ d \phi_{} } = a \sec ^{2} \left( \phi_{} \right) $$
$$ \frac{ dx }{ d \phi_{} } = \frac{ a }{ \cos ^{2} \left(\phi_{} \right) } $$
$$ dx \cdot \cos ^{2} \left(\phi_{} \right)= a \cdot d \phi_{} $$
$$ dx \cdot \left( \cos\left(\phi_{} \right) \right) ^{2} =a \cdot d \phi_{} $$
$$ dx \cdot \left( \cos\left(\frac{\pi}{2} -\theta_{} \right) \right) ^{2} =a \cdot d \phi_{} $$
$$ dx \cdot \left( \sin\left(\theta_{} \right) \right) ^{2} =a \cdot d \phi_{} $$
$$ dx \cdot \left( \frac{ a }{ \sqrt{ x ^{2} + a ^{2} } } \right) ^{2} = a \cdot d \phi_{} $$
$$ dx \cdot \left( \frac{ a ^{2} }{ x ^{2} + a ^{2} } \right) = a \cdot d \phi_{} $$
$$ dx \cdot \frac{ a }{ x ^{2} + a ^{2} } = d \phi_{} $$
I've been stucked from here.
$$ x= a \tan \left( \phi_{} \right) ~~ \leftarrow~~ \text{ok} $$
Just differentiate the formula with $~\phi~$ to obtain the stuff.