# Question #d1324

Aug 23, 2017

#### Explanation:

Enjoy trigonometry

Aug 24, 2017

See explanation below.

#### Explanation:

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

We will begin the proof from the right-hand side of the identity.

First, let's expand the parentheses:

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

Then, let's apply the standard trigonometric identities $\sec \left(x\right) = \frac{1}{\cos \left(x\right)}$ and $\tan \left(x\right) = \frac{\sin \left(x\right)}{\cos \left(x\right)}$:

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

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

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

Let's factorise the numerator of the fraction using the "middle-term break":

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

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

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

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

One of the Pythagorean identities is ${\cos}^{2} \left(x\right) + {\sin}^{2} \left(x\right) = 1$.

We can rearrange it to get:

$R i g h t a r r o w {\cos}^{2} \left(x\right) = 1 - {\sin}^{2} \left(x\right)$

Let's apply this rearranged identity to our proof:

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

Let's factorise the denominator, which is in the form of a difference of two squares:

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

$= \frac{1 + \sin \left(x\right)}{1 - \sin \left(x\right)} \text{ " }$ $\text{Q.E.D.}$