# How do you prove (1 - sin x)/(1 + sin x)=(sec x + tan x)^2?

Jul 18, 2016

Use a few trig identities and simplify. See below.

#### Explanation:

I believe there is a mistake in the question, but it's no big deal. In order for it to make sense, the question should read:
$\frac{1 - \sin x}{1 + \sin x} = {\left(\sec x - \tan x\right)}^{2}$

Either way, we start with this expression:
$\frac{1 - \sin x}{1 + \sin x}$

(When proving trig identities, it is generally best to work on the side that has a fraction).

Let's use a neat trick called conjugate multiplication, where we multiply the fraction by the denominator's conjugate:
$\frac{1 - \sin x}{1 + \sin x} \cdot \frac{1 - \sin x}{1 - \sin x}$

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

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

The conjugate of $a + b$ is $a - b$, so the conjugate of $1 + \sin x$ is $1 - \sin x$; we multiply by $\frac{1 - \sin x}{1 - \sin x}$ to balance the fraction.

Note that $\left(1 + \sin x\right) \left(1 - \sin x\right)$ is actually a difference of squares, which has the property:
$\left(a - b\right) \left(a + b\right) = {a}^{2} - {b}^{2}$

Here, we see that $a = 1$ and $b = \sin x$, so:
$\left(1 + \sin x\right) \left(1 - \sin x\right) = {\left(1\right)}^{2} - {\left(\sin x\right)}^{2} = 1 - {\sin}^{2} x$

From the Pythagorean Identity ${\sin}^{2} x + {\cos}^{2} x = 1$, it follows that (after subtracting ${\sin}^{2} x$ from both sides), ${\cos}^{2} x = 1 - {\sin}^{2} x$.

Wow, we went from $\frac{1 - \sin x}{1 - \sin x}$ to $1 - {\sin}^{2} x$ to ${\cos}^{2} x$! Now our problem looks like:
${\left(1 - \sin x\right)}^{2} / {\cos}^{2} x = {\left(\sec x - \tan x\right)}^{2}$

Let's expand the numerator:
$\frac{1 - 2 \sin x + {\sin}^{2} x}{\cos} ^ 2 x = {\left(\sec x - \tan x\right)}^{2}$
(Remember: ${\left(a - b\right)}^{2} = {a}^{2} - 2 a b + {b}^{2}$)

Now, we'll break up the fractions:
$\frac{1}{\cos} ^ 2 x - \frac{2 \sin x}{\cos} ^ 2 x + {\sin}^{2} \frac{x}{\cos} ^ 2 x$

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

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

How to simplify that? Well, remember when I said "Remember: ${\left(a - b\right)}^{2} = {a}^{2} - 2 a b + {b}^{2}$"?

It turns out that ${\sec}^{2} x - 2 \tan x \sec x + {\tan}^{2} x$ is actually ${\left(\sec x - \tan x\right)}^{2}$. If we let $a = \sec x$ and $b = \tan x$, we can see that this expression is:
${\underbrace{{\left(a\right)}^{2}}}_{\sec} x - 2 \left(a\right) \left(b\right) + {\underbrace{{\left(b\right)}^{2}}}_{\tan} x$

Which, as I just said is equivalent to ${\left(a - b\right)}^{2}$. Replace $a$ with $\sec x$ and $b$ with $\tan x$ and you get:
${\sec}^{2} x - 2 \tan x \sec x + {\tan}^{2} x = {\left(\sec x - \tan x\right)}^{2}$

And we have completed the prood:
${\left(\sec x - \tan x\right)}^{2} = {\left(\sec x - \tan x\right)}^{2}$