Garrett Black

2022-06-20

The identity $\mathrm{tan}\left(\frac{\pi }{4}-\frac{a}{2}\right)=\mathrm{sec}\left(a\right)-\mathrm{tan}\left(a\right)$

lorienoldf7

Denote $\mathrm{tan}\left(\alpha /2\right)=t$. Then
$\mathrm{sin}\alpha =\frac{2t}{1+{t}^{2}},\phantom{\rule{1em}{0ex}}\mathrm{cos}\alpha =\frac{1-{t}^{2}}{1+{t}^{2}},\phantom{\rule{1em}{0ex}}\mathrm{tan}\alpha =\frac{2t}{1-{t}^{2}}.$
Then
$\mathrm{tan}\left(\frac{\pi }{4}-\frac{\alpha }{2}\right)=\frac{1-t}{1+t}=\frac{\left(1-t\right)\left(1-t\right)}{\left(1+t\right)\left(1-t\right)}=\frac{\left(1+{t}^{2}\right)-2t}{1-{t}^{2}}$
$=\frac{1+{t}^{2}}{1-{t}^{2}}-\frac{2t}{1-{t}^{2}}=\frac{1}{\mathrm{cos}\alpha }-\mathrm{tan}\alpha$
$=\mathrm{sec}\alpha -\mathrm{tan}\alpha .$

telegrafyx

The identity
$\mathrm{sec}x±\mathrm{tan}x=\mathrm{tan}\left(\frac{\pi }{4}±\frac{x}{2}\right)$
is one that I sometimes think of as the "cartographers' tangent half-angle formula" because of the way in which it arises in the theory of the Mercator projection, i.e. in finding the antiderivative of the secant function. The desire to find the antiderivative of the secant function came from its application to cartography in the late 1500s and early 1600s.
Recall that $\mathrm{sec}\left(±x\right)=\mathrm{sec}x$ and $\mathrm{tan}\left(±x\right)=±\mathrm{tan}x$, so when the sign of x changes, the sign of $\mathrm{tan}x$ changes but that of $\mathrm{sec}x$ does not. Or in other words, secant is even and tangent is odd.
Since x appears on the right side only in $\frac{x}{2}$, and on the right side without such a fraction, you should expect to either prove a half-angle formula or rely on one that's already proved. Recall that
$\mathrm{tan}\frac{x}{2}=\frac{\mathrm{sin}x}{1+\mathrm{cos}x}.$
So
$\begin{array}{rl}& \mathrm{tan}\left(\frac{\pi }{4}±\frac{x}{2}\right)=\frac{\mathrm{sin}\left(\frac{\pi }{2}±x\right)}{1+\mathrm{cos}\left(\frac{\pi }{2}±x\right)}=\frac{\mathrm{cos}x}{1\mp \mathrm{sin}x}=\frac{\left(\mathrm{cos}x\right)\left(1±\mathrm{sin}x\right)}{\left(1\mp \mathrm{sin}x\right)\left(1±\mathrm{sin}x\right)}\\ =& \frac{\left(\mathrm{cos}x\right)\left(1±\mathrm{sin}x\right)}{1-{\mathrm{sin}}^{2}x}=\frac{\left(\mathrm{cos}x\right)\left(1±{\mathrm{sin}}^{2}x\right)}{{\mathrm{cos}}^{2}x}=\frac{1±\mathrm{sin}x}{\mathrm{cos}x}=\frac{1}{\mathrm{cos}x}±\frac{\mathrm{sin}x}{\mathrm{cos}x}\\ =& \mathrm{sec}x±\mathrm{tan}x.\end{array}$

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