How do you prove that (1 + tan(x/2)) / (1 - tan(x/2))= tan x + sec x?

Apr 16, 2015

The easier and faster way is to use the Tangent half-angle substitution formulae:

$\tan \left(\frac{x}{2}\right) = t$

$\tan x = \frac{2 t}{1 - {t}^{2}}$

$\cos x = \frac{1 - {t}^{2}}{1 + {t}^{2}}$. (remember that $\sec x = \frac{1}{\cos} x$)

So:

$\frac{1 + t}{1 - t} = \frac{2 t}{1 - {t}^{2}} + \frac{1 + {t}^{2}}{1 - {t}^{2}}$

and the second member becomes:

$\frac{2 t + 1 + {t}^{2}}{1 - {t}^{2}} = {\left(1 + t\right)}^{2} / \left(\left(1 - t\right) \left(1 + t\right)\right) \Rightarrow \frac{1 + t}{1 - t}$ that is the first member.