Landen Barber

2022-03-29

Prove that $\mathrm{tan}\left[\frac{\pi }{4}+\frac{1}{2}{\mathrm{cos}}^{-1}\left(\frac{a}{b}\right)\right]+\mathrm{tan}\left[\frac{\pi }{4}-\frac{1}{2}{\mathrm{cos}}^{-1}\left(\frac{a}{b}\right)\right]=\frac{2b}{a}$

Malia Booth

Using
$\mathrm{tan}\left(A+B\right)=\frac{\mathrm{tan}A+\mathrm{tan}B}{1-\mathrm{tan}A\mathrm{tan}B}$
There are
$\mathrm{tan}\left[\frac{\pi }{4}±\frac{1}{2}{\mathrm{cos}}^{-1}\frac{a}{b}\right]=\frac{1±\mathrm{tan}\frac{1}{2}{\mathrm{cos}}^{-1}\frac{a}{b}}{1\mp \mathrm{tan}\frac{1}{2}{\mathrm{cos}}^{-1}\frac{a}{b}}$
Hence
$\mathrm{tan}\left[\frac{\pi }{4}+\frac{1}{2}{\mathrm{cos}}^{-1}\left(\frac{a}{b}\right)\right]+\mathrm{tan}\left[\frac{\pi }{4}-\frac{1}{2}{\mathrm{cos}}^{-1}\left(\frac{a}{b}\right)\right]=2\frac{1+{\mathrm{tan}}^{2}\left(\frac{1}{2}{\mathrm{cos}}^{-1}\frac{a}{b}\right)}{1-{\mathrm{tan}}^{2}\left(\frac{1}{2}{\mathrm{cos}}^{-1}\frac{a}{b}\right)}$
using the half-angle formula, proceed.
$\mathrm{cos}\alpha =\frac{1-{\mathrm{tan}}^{2}\frac{\alpha }{2}}{1+{\mathrm{tan}}^{2}\frac{\alpha }{2}}$
There are
$\mathrm{tan}\left[\frac{\pi }{4}+{\frac{12}{\mathrm{cos}}}^{-1}\left(\frac{a}{b}\right)\right]+\mathrm{tan}\left[\frac{\pi }{4}-\frac{1}{2}{\mathrm{cos}}^{-1}\left(\frac{a}{b}\right)\right]=\frac{2}{{\mathrm{cos}\mathrm{cos}}^{-1}\frac{a}{b}}=\frac{2b}{a}$

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