Trignometric rearrangement

Question

So there is this equation:

$y=\tan \left(e^{x}+c\right)$

And in the next step it's differentiated w.r.t x

$\frac{d y}{d x}=\sec ^{2}\left(e^{x}+c\right) \cdot e^{x}$

Upto this point I follow what is going on.

And then apparently it is rearranged but the author didn't make any comment as to how he rearranged it as follows:

$\frac{d y}{d x}=\left(1+\tan ^{2}\left(e^{x}+c\right)\right) e^{x}$

$\frac{d y}{d x}= e^{x}+e^{x} y^{2}$

I can't figure out exactly how this is done, so I would appreciate any insights into this. Couldn't think of a better title so, feel free to change it if you want.

Answer 1

First is a trigonometric identity: $$ 1 + \tan^2 \theta = \sec^2 \theta $$ for any $\theta$. This can be proven by taking $\sin^2\theta + \cos^2\theta =1$ and dividing every term by $\cos^2\theta$.

Second, $1 + \tan^2(e^x + c)$ is swapped out for $1+y^2$, and that term is multiplied by $e^x$.

Answer 2

It's really simple: $$1+\tan^2 x=\frac{\cos^2x+\sin^2x}{\cos^2x}=\frac{1}{\cos^2x}=\sec^2x$$