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

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

\[ \tan x+\cot x=\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x} =\frac{\sin^2x+\cos^2x}{\sin x\cos x} =\frac{1}{\sin x\cos x} \Rightarrow \frac{1}{\tan x+\cot x}=\sin x\cos x. \]

\[ \sec^2x+\sec x\tan x=\frac{1}{\cos^2x}+\frac{\sin x}{\cos^2x} =\frac{1+\sin x}{\cos^2x} =\frac{1+\sin x}{1-\sin^2x} =\frac{1}{1-\sin x}. \]

\[ \frac{1-\cos^2x}{\sec^2x-1}=\frac{\sin^2x}{\tan^2x} =\frac{\sin^2x}{\frac{\sin^2x}{\cos^2x}}=\cos^2x=1-\sin^2x. \]