Identity to verify:
[tex]\dfrac{\cot x}{1+\csc x}=\dfrac{\csc x-1}{\cot x}[/tex]
Recall that
[tex]\cos^2x+\sin^2x=1[/tex]
Divide both sides by [tex]\sin^2x[/tex] and we get
[tex]\cot^2x+1=\csc^2x[/tex]
or
[tex]\cot^2x=\csc^2x-1=(\csc x-1)(\csc x+1)[/tex]
So if we multiply the numerator and denominator of
[tex]\dfrac{\cot x}{1+\csc x}[/tex]
by [tex]\csc x-1[/tex], we get
[tex]\dfrac{\cot x(\csc x-1)}{(1+\csc x)(\csc x-1)}=\dfrac{\cot x(\csc x-1)}{\csc^2x-1}=\dfrac{\cot x(\csc x-1)}{\cot^2x}[/tex]
Then as long as [tex]\cot x\neq0[/tex], we can cancel terms to end up with
[tex]\dfrac{\csc x-1}{\cot x}[/tex]
and establish the identity.