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(Maybe I'll post my own answer here, but maybe others will make that redundant.)

This is a fun (?) trivia item that fell out of a bit of geometry I was thinking about.

One of the tangent half-angle formulas says $$ \tan\frac\alpha2 = \frac{\sin\alpha}{1+\cos\alpha}. $$

Consider the quadratic equation $$ x^2 + 2bx - 1 = 0. $$ Solving for $b$, we get $$ b = \frac{1-x^2}{2x}. $$ Let $\alpha=\arctan b$. Since $\tan\alpha=(1-x^2)/2x$, we have $\sin\alpha=(1-x^2)/(1+x^2)$ and $\cos\alpha=2x/(1-x^2)$, so $$ \tan\frac\alpha2=\frac{1-x^2}{(1+x)^2}. $$ Lo and behold, this fraction is not in lowest terms, and we have $$ \tan\frac\alpha2= \frac{\sin\alpha}{1+\cos\alpha} = \frac{1-x}{1+x}. $$ This implies $$ (1+x)\tan\frac\alpha2= 1-x $$ and so we seem to have reduced a second-degree equation to a first-degree equation by using the tangent half-angle formula. Solving for $x$, we get $$ x=\frac{1-\tan\frac\alpha2}{1+\tan\frac\alpha2}.\tag{1} $$ Since $\tan\alpha=b$, we have $\sin\alpha=b/\sqrt{1+b^2}$ and $\cos\alpha=1/\sqrt{1+b^2}$, so $$ \tan\frac\alpha2=\frac{\sin\alpha}{1+\cos\alpha}=\frac{b}{1+\sqrt{1+b^2}}. $$ Sustituting this back into $(1)$, we get $$ x = \frac{1+\sqrt{b^2-1}-b}{1+\sqrt{b^2+1}+b} = -b+\sqrt{b^2+1}. $$ This is one of the two solutions of the quadratic equation. What happened to the other one?

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    Just maybe, the form of the tangent half-angle formula most convenient for the geometry problem (whose content is not stated here) will be the one used in cartography: $\tan\left(\dfrac\alpha2+\dfrac\pi4\right)=\sec\alpha+\tan\alpha.\qquad{}$2012-12-07

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