# Question #e51b8

Jan 22, 2017

$\tan \left(\frac{x}{2} + \frac{\pi}{4}\right) = \sec x + \tan x$

#### Explanation:

Use the trig identity:
$\tan \left(a + b\right) = \frac{\tan a + \tan b}{1 - \tan a . \tan b}$
Call t = (x/2) and develop the left side
$L S = \tan \left(t + \frac{\pi}{4}\right) = \frac{\tan t + \tan \left(\frac{\pi}{4}\right)}{1 - \tan t . \tan \left(\frac{\pi}{4}\right)}$
Trig table gives $\tan \left(\frac{\pi}{4}\right) = 1$, there for:
$L S = \frac{1 + \tan t}{1 - \tan t} = \left(\frac{\cos t + \sin t}{\cos t}\right) \left(\cos \frac{t}{\cos t - \sin t}\right) =$
$L S = \frac{\cos t + \sin t}{\cos t - \sin t}$
Multiply both numerator and denominator by (cos t + sin t), we get:
$L S = {\left(\cos t + \sin t\right)}^{2} / \left({\cos}^{2} t - {\sin}^{2} t\right)$
Reminder:
${\left(\cos t + \sin t\right)}^{2} = 1 + 2 \cos t . \sin t = 1 + \sin 2 t$
${\cos}^{2} t - {\sin}^{2} t = \cos 2 t$.
$L S = \frac{1 + \sin 2 t}{\cos 2 t} = \frac{1}{\cos 2 t} + \frac{\sin 2 t}{\cos 2 t}$
$L S = \sec 2 t + \tan 2 t$
Replace t by $\left(\frac{x}{2}\right)$, we get
$\tan \left(\frac{x}{2} + \frac{\pi}{4}\right) = \sec x + \tan x$

Jan 22, 2017

$R H S = \sec x + \tan x$

$= \frac{1}{\cos} x + \sin \frac{x}{\cos} x$

$= \frac{1 + \sin x}{\cos} x$

$= \frac{{\cos}^{2} \left(\frac{x}{2}\right) + {\sin}^{2} \left(\frac{x}{2}\right) + 2 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)}{{\cos}^{2} \left(\frac{x}{2}\right) - {\sin}^{2} \left(\frac{x}{2}\right)}$

$= {\left(\cos \left(\frac{x}{2}\right) + \sin \left(\frac{x}{2}\right)\right)}^{2} / \left(\left(\cos \left(\frac{x}{2}\right) + \sin \left(\frac{x}{2}\right)\right) \left(\cos \left(\frac{x}{2}\right) - \sin \left(\frac{x}{2}\right)\right)\right)$

$= \frac{\cos \left(\frac{x}{2}\right) + \sin \left(\frac{x}{2}\right)}{\cos \left(\frac{x}{2}\right) - \sin \left(\frac{x}{2}\right)}$

$= \frac{\cos \frac{\frac{x}{2}}{\cos} \left(\frac{x}{2}\right) + \sin \frac{\frac{x}{2}}{\cos} \left(\frac{x}{2}\right)}{\cos \frac{\frac{x}{2}}{\cos} \left(\frac{x}{2}\right) - \sin \frac{\frac{x}{2}}{\cos} \left(\frac{x}{2}\right)}$

$= \frac{1 + \tan \left(\frac{x}{2}\right)}{1 - \tan \left(\frac{x}{2}\right)}$

$= \frac{\tan \left(\frac{\pi}{4}\right) + \tan \left(\frac{x}{2}\right)}{1 - \tan \left(\frac{x}{2}\right) \tan \left(\frac{\pi}{4}\right)}$

$= \tan \left(\frac{x}{2} + \frac{\pi}{4}\right) = L H S$

Jan 22, 2017

Proof given below

#### Explanation:

$\tan \left(\frac{x}{2} + \frac{\pi}{4}\right) = \frac{\tan \frac{\pi}{4} + \tan \frac{x}{2}}{1 - \tan \frac{x}{2} \tan \frac{\pi}{4}}$

=$\frac{1 + \tan \frac{x}{2}}{1 - \tan \frac{x}{2}}$ = $\frac{\cos \frac{x}{2} + \sin \frac{x}{2}}{\cos \frac{x}{2} - \sin \frac{x}{2}}$

Now multiply the numerator and denominator by $\left(\cos \frac{x}{2} + \sin \frac{x}{2}\right)$

= $\frac{1 + 2 \sin \frac{x}{2} \cos \frac{x}{2}}{{\cos}^{2} \frac{x}{2} - {\sin}^{2} \frac{x}{2}}$ = $\frac{1 + \sin x}{\cos} x$= sec x + tanx