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

Jul 10, 2016

Do some conjugate multiplication, make use of trig identities, and simplify. See below.

Explanation:

Recall the Pythagorean Identity ${\sin}^{2} x + {\cos}^{2} x = 1$. Divide both sides by ${\cos}^{2} x$:
$\frac{{\sin}^{2} x + {\cos}^{2} x}{\cos} ^ 2 x = \frac{1}{\cos} ^ 2 x$
$\to {\tan}^{2} x + 1 = {\sec}^{2} x$
We will be making use of this important identity.

Let's focus on this expression:
$\sec x + 1$

Note that this is equivalent to $\frac{\sec x + 1}{1}$. Multiply the top and bottom by $\sec x - 1$ (this technique is known as conjugate multiplication):
$\frac{\sec x + 1}{1} \cdot \frac{\sec x - 1}{\sec x - 1}$
$\to \frac{\left(\sec x + 1\right) \left(\sec x - 1\right)}{\sec x - 1}$
$\to \frac{{\sec}^{2} x - 1}{\sec x - 1}$

From ${\tan}^{2} x + 1 = {\sec}^{2} x$, we see that ${\tan}^{2} x = {\sec}^{2} x - 1$. Therefore, we can replace the numerator with ${\tan}^{2} x$:
$\frac{{\tan}^{2} x}{\sec x - 1}$

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

We have a common denominator, so we can add the fractions on the left hand side:
$\frac{{\tan}^{2} x}{\sec x - 1} + \frac{1 - {\tan}^{2} x}{\sec x - 1} = \cos \frac{x}{1 - \cos x}$
$\to \frac{{\tan}^{2} x + 1 - {\tan}^{2} x}{\sec x - 1} = \cos \frac{x}{1 - \cos x}$

The tangents cancel:
$\frac{\cancel{{\tan}^{2} x} + 1 - \cancel{{\tan}^{2} x}}{\sec x - 1} = \cos \frac{x}{1 - \cos x}$

Leaving us with:
$\frac{1}{\sec x - 1} = \cos \frac{x}{1 - \cos x}$

Since $\sec x = \frac{1}{\cos} x$, we can rewrite this as:
$\frac{1}{\frac{1}{\cos} x - 1} = \cos \frac{x}{1 - \cos x}$

Adding fractions in the denominator, we see:
$\frac{1}{\frac{1}{\cos} x - 1} = \cos \frac{x}{1 - \cos x}$
$\to \frac{1}{\frac{1}{\cos} x - \frac{\cos x}{\cos x}} = \cos \frac{x}{1 - \cos x}$

$\to \frac{1}{\frac{1 - \cos x}{\cos} x} = \cos \frac{x}{1 - \cos x}$

Using the property $\frac{1}{\frac{a}{b}} = \frac{b}{a}$, we have:
$\cos \frac{x}{1 - \cos x} = \cos \frac{x}{1 - \cos x}$

And that completes the proof.

Jul 11, 2016

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

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

$= \frac{{\sec}^{2} x - 1 + 1 - {\tan}^{2} x}{\sec x - 1}$

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

$\textcolor{red}{\text{putting} , {\sec}^{2} x - {\tan}^{2} x = 1}$

$= \cos \frac{x}{\cos x \sec x - \cos x}$

$\textcolor{red}{\text{putting} , \cos x \sec x = 1}$

$= \cos \frac{x}{1 - \cos x} = R H S$