Why is $\tan^2(y)=x^2$?

Question

I'm following a proof in a book about the derivative of an arc tangent as follows:

By definition, $y = \tan^{-1}(x) \Rightarrow \tan(y)=x$

Therefore, using implicit derivation:

$$ \sec^2(y)\frac{dy}{dx}=1 \\ \frac{dy}{dx}=\frac{1}{\sec^2(y)}=\frac{1}{1+\tan^2(y)}=\frac{1}{1+x^2} $$

Thus completing the proof. I follow well the implicit derivation, and the trigonometric identity $\sec^2(x)=1+\tan^2(x)$, but I don't follow how did it jump to the conclusion that $\tan^2(y)=x^2$. I can't find a reference for this.

What would be the reason for this? Thanks.

Answer 1

Simply from here

$$y = \tan^{-1}(x) \Rightarrow \tan(y) = \tan(\tan^{-1}(x)) \Rightarrow\tan(y)=x\Rightarrow (\tan(y))^2=x^2$$

Answer 2

This results from the very definition of $\arctan $: $$y=\arctan x\overset{\text{def.}}\iff \tan y=x\enspace\textbf{and}\enspace -\tfrac\pi2<y<\tfrac\pi2.$$