# Prove that (1+sinx)/(1-sinx)=(secx+tanx)^2?

Oct 13, 2016

Using the identities

• ${\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right) = 1 \implies 1 - {\sin}^{2} \left(x\right) = {\cos}^{2} \left(x\right)$
• $\left(a - b\right) \left(a + b\right) = {a}^{2} - {b}^{2}$

we have

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

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

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

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

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

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

Oct 13, 2016

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

#### Explanation:

Let us start from right hand side.

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

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

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

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

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

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

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

Oct 13, 2016

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

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

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

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

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

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

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

Proved