# How do you prove Tan^2(x/2+Pi/4)=(1+sinx)/(1-sinx)?

Apr 23, 2016

Proof below (it's a long one)

#### Explanation:

Ill work this backwards (but writing doing it forward would work as well):
$\frac{1 + \sin x}{1 - \sin x} = \frac{1 + \sin x}{1 - \sin x} \cdot \frac{1 + \sin x}{1 + \sin x}$
$= {\left(1 + \sin x\right)}^{2} / \left(1 - {\sin}^{2} x\right)$
$= {\left(1 + \sin x\right)}^{2} / {\cos}^{2} x$
$= {\left(\frac{1 + \sin x}{\cos} x\right)}^{2}$
Then substitute in $t$ formula (Explanation below)
$= {\left(\frac{1 + \frac{2 t}{1 + {t}^{2}}}{\frac{1 - {t}^{2}}{1 + {t}^{2}}}\right)}^{2}$
$= {\left(\frac{\frac{1 + {t}^{2} + 2 t}{1 + {t}^{2}}}{\frac{1 - {t}^{2}}{1 + {t}^{2}}}\right)}^{2}$
$= {\left(\frac{1 + {t}^{2} + 2 t}{1 - {t}^{2}}\right)}^{2}$
$= {\left(\frac{1 + 2 t + {t}^{2}}{1 - {t}^{2}}\right)}^{2}$
$= {\left({\left(1 + t\right)}^{2} / \left(1 - {t}^{2}\right)\right)}^{2}$
$= {\left({\left(1 + t\right)}^{2} / \left(\left(1 - t\right) \left(1 + t\right)\right)\right)}^{2}$
$= {\left(\frac{1 + t}{1 - t}\right)}^{2}$
$= {\left(\frac{1 + \tan \left(\frac{x}{2}\right)}{1 - \tan \left(\frac{x}{2}\right)}\right)}^{2}$
$= {\left(\frac{\tan \left(\frac{\pi}{4}\right) + \tan \left(\frac{x}{2}\right)}{1 - \tan \left(\frac{x}{2}\right) \tan \left(\frac{\pi}{4}\right)}\right)}^{2}$ Note that: (tan(pi/4)=1)
$= {\left(\tan \left(\frac{x}{2} + \frac{\pi}{4}\right)\right)}^{2}$
$= {\tan}^{2} \left(\frac{x}{2} + \frac{\pi}{4}\right)$

T FORMULAE FOR THIS EQUATION:
$\sin x = \frac{2 t}{1 - {t}^{2}}$, $\cos x = \frac{1 - {t}^{2}}{1 + {t}^{2}}$, where $t = \tan \left(\frac{x}{2}\right)$