# Question #5060b

Mar 1, 2017

Use the trigonometric identity that states ${\sin}^{2} \theta + {\cos}^{2} \theta = 1$

#### Explanation:

Typically, a mathematical proof follows a series of logical arguments to show that some theorem is true based on a set of axioms (one or more basic concepts that are assumed to be true).

If one can reach a logical conclusion based on an axiom, without committing any mathematical errors, then the theorem is "proven".

Let's start by assuming the following is true:

${\sin}^{2} \theta + {\cos}^{2} \theta = 1$

Subtracting ${\sin}^{2} \theta$ from both sides, we get:

${\cos}^{2} \theta = 1 - {\sin}^{2} \theta$

Let's replace $\cos \theta$ with $u$ and $\sin \theta$ with $v$ for a moment:

${u}^{2} = 1 - {v}^{2}$

Looking at the right hand side of the equation, it is a difference of squares, and so it can be factored into the following:

${u}^{2} = \left(1 - v\right) \left(1 + v\right)$

Now, let's divide both sides by $\left(1 - v\right)$

${u}^{2} / \left(1 - v\right) = \left(1 + v\right)$

And, now, let's divide both sides by $u$

$\frac{u}{1 - v} = \frac{1 + v}{u}$

Let's separate the terms on the right hand side of the equation:

$\frac{u}{1 - v} = \frac{1}{u} + \frac{v}{u}$

Finally, let's replace $u$ and $v$ with $\cos \theta$ and $\sin \theta$:

$\cos \frac{\theta}{1 - \sin \theta} = \frac{1}{\cos} \theta + \sin \frac{\theta}{\cos} \theta$

Simplifying, we get:

$\cos \frac{\theta}{1 - \sin \theta} = \sec \theta + \tan \theta$

Since we were able to logically show that the end result was true based on our initial assumption which was known to be true, it must be the case that $\cos \frac{\theta}{1 - \sin \theta} = \sec \theta + \tan \theta$

Note: This problem can also be worked in reverse!