Proving

use the identity (a+b)(a-b) = a^2  - b^2  with {(sec x)-1}{ (sec x +1) }   to get

sec^2(x)  -  1

we can now write expression as

{sec^2(x) -1} * cot (x)    

and use the trig identity  sec^2(x) - 1   = tan^2(x)    to get expression as

tan^2(x)*cot(x)   = tan^2(x) * {1/tan(x)}     = tan(x)

$\text{LHS}\\ =(\sec x - 1)(\sec x + 1)(\cot x) \\ =(\sec^2 x - 1)(\cot x)\\ =(\tan^2 x)(\cot x)\\ =\tan x\\ =\text{RHS}$