# How do you prove (1+sinx)/(1-sinx)=(secx+tanx)^2?

Oct 3, 2016

see below

#### Explanation:

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

Right Side $= {\left(\sec x + \tan x\right)}^{2}$

$= \left(\sec x + \tan x\right) \left(\sec x + \tan x\right)$

$= {\sec}^{2} x + 2 \sec x \tan x + {\tan}^{2} x$

$= \frac{1}{\cos} ^ 2 x + 2 \cdot \frac{1}{\cos} x \cdot \sin \frac{x}{\cos} x + {\sin}^{2} \frac{x}{\cos} ^ 2 x$

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

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

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

=(1+sinx)/(1-sinx

$=$ Left Side

Feb 20, 2018

A simpler derivation is given below :

#### Explanation:

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