Respuesta :

I guess you mean

[tex]\sec^6x(\sec x\tan x)-\sec^4x(\sec x\tan x)=\sec^5x\tan^3x[/tex]

On the left side, we have a common factor of [tex]\sec^4x(\sec x\tan x)=\sec^5x\tan x[/tex], so that

[tex]\sec^6x(\sec x\tan x)-\sec^4x(\sec x\tan x)=\sec^5x\tan x(\sec^2x-1)[/tex]

Recall that

[tex]\sec^2x=1+\tan^2x[/tex]

from which it follows that

[tex]\sec^5x\tan x(\sec^2x-1)=\sec^5x\tan x\tan^2x=\sec^5x\tan^3x[/tex]