# How do you prove this identity?

How do you simplify this using the Reciprocal, Cofunction, and Pythagorean Identities?

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

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I believe you will find it easiest to start on the RHS of the equation. Rewrite the $\sec x$ as $\frac{1}{\cos x}$

$$\frac{\frac{1}{\cos x} + 1}{\frac{1}{\cos x} - 1}$$

Now you have a complex fraction. So you multiply both top and bottom by your common denominator. After simplifying you should be there.

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Starting from the LHS:

$\dfrac{1 + \cos(x)}{1 - \cos(x)}$

Factor out cos(x) from top and bottom as follows:

$\dfrac{\cos(x)\left(\dfrac{1}{\cos(x)} + 1\right)}{\cos(x)\left(\dfrac{1}{\cos(x)} - 1\right)}$

The factored $\cos(x)$ on top and bottom cancels leaving just:

$\dfrac{\dfrac{1}{\cos(x)} + 1}{\dfrac{1}{\cos(x)} - 1}$

From here, it is easy to see that since $\dfrac{1}{\cos(x)} = \sec(x)$:

$\dfrac{\sec(x) + 1}{\sec(x) - 1}$,

Ending with the RHS.

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