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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: $\tan\left(\dfrac\alpha2+\dfrac\pi4\right)=\sec\alpha+\tan\alpha.\qquad{}$ –  Michael Hardy Dec 7 '12 at 18:32
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I've up-voted marty cohen's answer, but I would add this:

Maybe the first omission could be said to be the use of $\arctan b$ instead of using both of the points on the circle corresponding to $\tan=b$. For the conventional value of $\arctan b$, the positive square root of $1+b^2$ is the only one that's right.

When you include that other angle whose tangent is $\alpha$, the sine and cosine of the angle are minus what they were before, and then the half-angle, and its tangent, are different from what they otherwise were, and we do get the other solution of the quadratic equation.

But strictly speaking the first actual error is in the sentence "Since $\tan\alpha=(1-x^2)/2x$, we have $\sin\alpha=(1-x^2)/(1+x^2)$ and $\cos\alpha=2x/(1-x^2)$. The fact is there are two points on the circle where $\tan=b$ and we need both of them.

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That is a much more through and accurate answer than my hasty one. Thanks for the upvote. –  marty cohen Dec 7 '12 at 6:36
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When you took $\sqrt{1+b^2}$ you did not consider the negative square root.

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