Prove: $\cos^2 x (\sec x - 1)(\sec x + 1) = (1 - \cos x)(1 + \cos x)$ 
Prove the trigonometric identity
  $$\cos^2 x (\sec x - 1)(\sec x + 1) = (1 - \cos x)(1 + \cos x)$$

I've searched high and low on the net and cannot find identities where there is $+$ or $- 1$'s in the equation. Any help is appreciated.    
Edit after reviewing comments.
$\cos^2 {x} (\sec {x} -1)(\sec {x} +1) = (1 - \cos {x})(1 + \cos {x})\quad$
$\cos^2 {x} (\sec^2 {x}- 1) = 1 - \cos^2 {x}\quad$
$\cos^2 {x}·tan^2 {x} = \sin^2 {x} \quad$
$\cos^2 {x} ·\frac{\sin^2 {x}}{\cos^2 {x}} = \sin^2 {x} \quad$
$\frac{\sin^2 {x}}{\cos^2 {x} }=\frac{\sin^2 {x}}{\cos^2 {x} }\quad$  
How's that? I tried the $\sec {x }= \frac1{\cos {x}}\quad$ method.    However I was unsuccessful.    

Substitute $\frac1{\cos{x}}\;$ for $\sec{x}\;$ in the identity, and it will shout its truth to you.
