\[ \csc x-\cot x=\frac{1-\cos x}{\sin x},\qquad 1+\sec x=1+\frac{1}{\cos x}=\frac{1+\cos x}{\cos x}. \] Thus \[ (1+\sec x)(\csc x-\cot x)= \frac{(1-\cos x)(1+\cos x)}{\sin x\cos x} =\frac{1-\cos^2x}{\sin x\cos x} =\frac{\sin^2x}{\sin x\cos x}=\tan x. \]