How do you verify the Identity: (1+cos x+ sinx)/(1+cos x-sin x)=sec x+tan x?

Mar 26, 2018

$\frac{1 + \cos x + \sin x}{1 + \cos x - \sin x} = \sec x + \tan x | \cdot \left(1 + \cos x - \sin x\right)$
$1 + \cos x + \sin x = \left(\frac{1}{\cos} x + \sin \frac{x}{\cos} x\right) \cdot \left(1 + \cos x - \sin x\right)$
$= \frac{1 + \cos x - \sin x}{\cos} x + \left(1 + \cos x - \sin x\right) \cdot \sin \frac{x}{\cos} x$

$\frac{1 + \cos x - \sin x}{\cos} x = \textcolor{b l u e}{\sec x + 1 - \tan x}$
$\left(1 + \cos x - \sin x\right) \cdot \sin \frac{x}{\cos} x = \textcolor{red}{\tan x + \sin x - \sin x \tan x}$

$1 + \cos x + \sin x = \textcolor{b l u e}{\sec x + 1 \cancel{- \tan x}} + \textcolor{red}{\cancel{\tan x} + \sin x - \sin x \tan x}$
$1 + \cos x + \sin x = \sec x + 1 + \sin x - \sin x \tan x | - 1 - \sin x$
$\cos x = \sec x - \sin x \tan x | \cdot \cos x$
${\cos}^{2} x = 1 - {\sin}^{2} x | + {\sin}^{2} x$
${\cos}^{2} x + {\sin}^{2} x = 1$

Mar 27, 2018

$L H S = \frac{1 + \cos x + \sin x}{1 + \cos x - \sin x}$

$= \frac{\cos x \left(1 + \cos x + \sin x\right)}{\cos x \left(1 + \cos x - \sin x\right)}$

$= \frac{\cos x \left(1 + \sin x\right) + {\cos}^{2} x}{\cos x \left(1 + \cos x - \sin x\right)}$

$= \frac{\cos x \left(1 + \sin x\right) + \left(1 - {\sin}^{2} x\right)}{\cos x \left(1 + \cos x - \sin x\right)}$

$= \frac{\cos x \left(1 + \sin x\right) + \left(1 - \sin x\right) \left(1 + \sin x\right)}{\cos x \left(1 + \cos x - \sin x\right)}$

$= \frac{\left(1 + \sin x\right) \left(\cos x + 1 - \sin x\right)}{\cos x \left(1 + \cos x - \sin x\right)}$

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

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

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