1. ## Simplifying with cosine

1-cos(^2)x/(cos(^2)x) - 1. How do I simplify this to sec(^2)x -2?

2. ## Re: Simplifying with cosine

I assume it's $\frac{1-\cos^2x}{\cos^2x}-1=\frac{1}{\cos^2x}-1-1$.

3. ## Re: Simplifying with cosine

How did you get from step one to 2? Yes, that is correct.

4. ## Re: Simplifying with cosine

What do you mean by steps 1 and 2?

For all a, b and c, $\frac{a-b}{c}=\frac{a}{c}-\frac{b}{c}$. Also, $\frac{\cos^2 x}{\cos^2 x}=1$ (if $\cos^2 x\ne0$).

5. ## Re: Simplifying with cosine

Hello, benny92000!

We will need a few identities:

. . $\sin^2\!x + \cos^2\!x \:=\:1 \quad\Rightarrow\quad 1-\cos^2\!x \:=\:\sin^2\!x$

. . $\tan x \:=\:\frac{\sin x}{\cos x}$

. . $\sec^2\!x - \tan^2\!x \:=\:1 \quad\Rightarrow\quad \tan^2\!x \:=\:\sec^2\!x-1$

$\text{Given: }\:\frac{1-\cos^2\!x}{\cos^2\!x} - 1$

$\text{How do I simplify this to: }\:\sec^2\!x -2$

$\frac{1-\cos^2\!x}{\cos^2\!x} - 1 \quad=\quad \frac{\sin^2\!x}{\cos^2\!x} - 1 \quad=\quad \left(\frac{\sin x}{\cos x}\right)^2 - 1$

. . . . $=\quad \tan^2\!x - 1 \quad=\quad (\sec^2\!x - 1) - 1 \quad=\quad \sec^2\!x-2$

6. ## Re: Simplifying with cosine

Why would we need those identities? To simplify the original expression one only needs the definition of sec.

7. ## Re: Simplifying with cosine

emakarov's approach is the way to go.

Alternatively

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

$=sec^2x-sec^2xcos^2x-1$

and again finish with

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