"Everybody knows" that \begin{align} \tan\theta & =\frac{2\tan\frac\theta2}{1 - \tan^2\frac\theta2} = \frac{2\sin\frac\theta2\cos\frac\theta2}{\cos^2\frac\theta2-\sin^2\frac\theta2} \tag 1 \\[10pt] & = \dfrac{\text{2 times a product}}{\text{a difference of two squares}}. \tag 2 \end{align}
In the course of thinking about a geometry problem, it became natural to write $$ \tan\dfrac\theta2=\dfrac{J-K}{J+K}. $$
Plugging this into $(1)$, we get \begin{align} \tan\theta & = \dfrac{J^2-K^2}{2JK} \\[10pt] & = \dfrac{\text{a difference of two squares}}{\text{2 times a product}}. \tag 3 \end{align}
So I am struck by
- the fact that the expression in $(3)$ looks like the reciprocal of that in $(2)$, although of course they are equal; and
- the thought that such a seemingly trivial thing is probably thought of by some people as something they've known since infancy and that they see repeatedly every day without giving it a thought, starting before breakfast and lasting until they're sipping their cognac in the evening.
So
- Is the second thought that strikes me correct; and
- In what contexts does this curious but trivially derived inversion arise?
To put it another way, is the inversion in some sense non-trivial even though its derivation is trivial?
PS: Here's a bit more about the context in which this came up. The circle $|z|=1$ in $\mathbb C$ is invariant under $z\mapsto w=g(z)=\dfrac{Jz+K}{Kz+J}$, for $J,K\in\mathbb R$, and that mapping has two fixed points: $\pm1$. The image of the imaginary unit $i$ under this mapping is therefore $e^{i\theta}$ for some $\theta\in\mathbb R/2\pi\mathbb Z$. It follows that $J=1+\tan\frac\theta2$ and $K=1-\tan\frac\theta2$. If we let $z=e^{i\alpha}$ and $w=g(z)=e^{i\beta}$ for some $\alpha,\beta\in\mathbb R/2\pi\mathbb Z$, then \begin{align} \tan\frac\beta2 & = \tan\frac\theta2\cdot\tan\frac\alpha2 \\[12pt] & = \frac{J-K}{J+K}\tan\frac\alpha2. \end{align} I've actually forgotten whether there was a reason why I wanted $\tan\theta$ in terms of $J$ and $K$. There is a yet broader context in which I had occasion to do this, but that would be quite off topic for the present question.
Edited: After I read the postscriptum, my initial response seemed not really to address the question. Here's a geometric interpretation.
Putting $t = \tan \frac{\theta}{2}$ and $\tau = K/J$, $$ t = \tan \frac{\theta}{2} = \frac{J - K}{J + K} = \frac{1 - \tau}{1 + \tau},\qquad \frac{2t}{1 - t^{2}} = \tan \theta = \frac{J^{2} - K^{2}}{2JK} = \frac{1 - \tau^{2}}{2\tau}. $$
Geometrically, the "reciprocal" relationship manifests the swapping of coordinate expressions on the circle in passing from $t$ to $\tau$, which in turn comes from the interpretations of $t$ and $\tau$ as affine coordinates on $\mathbf{R}$ with respect to stereographic projections from $(-1, 0)$ and $(0, -1)$, respectively.