# How do you change sec^3(x) - sec(x) into an expression with only sines and cosines?

## Use fundamental identities to change the expression sec^3(x) - sec(x) to one involving only sines and cosines. Then simplify. I am at a complete loss on how I can do this. I have spent an hour looking up fundamental identities and derivatives and I still can't make any progress on this problem at all.

May 10, 2017

Use $\sec \left(x\right) = \frac{1}{\cos} \left(x\right)$ and the Pythagorean identity ${\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right) = 1$

#### Explanation:

Formatted problem: ${\sec}^{3} \left(x\right) - \sec \left(x\right)$

Remember that $\sec \left(x\right) = \frac{1}{\cos} \left(x\right)$.

First, we can use the distributive property to take out a $\sec \left(x\right)$:
$\sec \left(x\right) \left({\sec}^{2} \left(x\right) - 1\right)$

Here, we consider the Pythagorean Identity ${\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right) = 1$, which can be modified by dividing both sides by ${\cos}^{2} \left(x\right)$ to give us:
${\sin}^{2} \frac{x}{\cos} ^ 2 \left(x\right) + {\cos}^{2} \frac{x}{\cos} ^ 2 \left(x\right) = \frac{1}{{\cos}^{2} \left(x\right)}$
which is equal to
${\tan}^{2} \left(x\right) + 1 = {\sec}^{2} \left(x\right)$
by subtracting $1$ from both sides, we get:
${\tan}^{2} \left(x\right) = {\sec}^{2} \left(x\right) - 1$ which is most applicable for this problem.

Continuing on, we see that we can substitute ${\tan}^{2} \left(x\right) = {\sec}^{2} \left(x\right) - 1$ into the problem:
$\sec \left(x\right) \left({\sec}^{2} \left(x\right) - 1\right)$
$= \sec \left(x\right) \left({\tan}^{2} \left(x\right)\right)$

Now, since we know that $\sec \left(x\right) = \frac{1}{\cos} \left(x\right)$ and $\tan \left(x\right) = \sin \frac{x}{\cos} \left(x\right)$, we can rewrite $\sec \left(x\right) \left({\tan}^{2} \left(x\right)\right)$ using only cosines and sines:
$\frac{1}{\cos} \left(x\right) \left({\sin}^{2} \frac{x}{\cos} ^ 2 \left(x\right)\right)$

By simplifying, we get our final answer as:
${\sin}^{2} \frac{x}{\cos} ^ 3 \left(x\right)$

Hope this helps.