1-cos(^2)x/(cos(^2)x) - 1. How do I simplify this to sec(^2)x -2?
Hello, benny92000!
We will need a few identities:
. . $\displaystyle \sin^2\!x + \cos^2\!x \:=\:1 \quad\Rightarrow\quad 1-\cos^2\!x \:=\:\sin^2\!x$
. . $\displaystyle \tan x \:=\:\frac{\sin x}{\cos x}$
. . $\displaystyle \sec^2\!x - \tan^2\!x \:=\:1 \quad\Rightarrow\quad \tan^2\!x \:=\:\sec^2\!x-1$
$\displaystyle \text{Given: }\:\frac{1-\cos^2\!x}{\cos^2\!x} - 1$
$\displaystyle \text{How do I simplify this to: }\:\sec^2\!x -2$
$\displaystyle \frac{1-\cos^2\!x}{\cos^2\!x} - 1 \quad=\quad \frac{\sin^2\!x}{\cos^2\!x} - 1 \quad=\quad \left(\frac{\sin x}{\cos x}\right)^2 - 1$
. . . . $\displaystyle =\quad \tan^2\!x - 1 \quad=\quad (\sec^2\!x - 1) - 1 \quad=\quad \sec^2\!x-2$
emakarov's approach is the way to go.
Alternatively
$\displaystyle \frac{1-cos^2x}{cos^2x}-1=\frac{1}{cos^2x}\left(1-cos^2x\right)-1=sec^2x\left(1-cos^2x\right)-1$
$\displaystyle =sec^2x-sec^2xcos^2x-1$
and again finish with
$\displaystyle cos^2x=\frac{1}{sec^2x}$