Answer:
[tex]\frac{\cot x}{1+\csc x}=\frac{\csc x-1}{\cot x}[/tex]
Step-by-step explanation:
We want to verify the identity:
[tex]\frac{\cot x}{1+\csc x}=\frac{\csc x-1}{\cot x}[/tex]
Let us take the LHS and simplify to get the LHS.
Express everything in terms of the cosine and sine function.
[tex]\frac{\cot x}{1+\csc x}=\frac{\frac{\cos x}{\sin x} }{1+\frac{1}{\sin x} }[/tex]
Collect LCM
[tex]\frac{\cot x}{1+\csc x}=\frac{\frac{\cos x}{\sin x} }{\frac{\sin x+1}{\sin x} }[/tex]
We simplify the RHS to get:
[tex]\frac{\cot x}{1+\csc x}=\frac{\cos x}{\sin x+1}[/tex]
We rationalize to get:
[tex]\frac{\cot x}{1+\csc x}=\frac{\cos x(\sin x-1)}{(\sin x+1)*(\sin x-1)}[/tex]
We expand to get:
[tex]\frac{\cot x}{1+\csc x}=\frac{\cos x(\sin x-1)}{\sin^2 x-1}[/tex]
Factor negative one in the denominator:
[tex]\frac{\cot x}{1+\csc x}=\frac{\cos x(\sin x-1)}{-(1-\sin^2 x)}[/tex]
Apply the Pythagoras Property to get:
[tex]\frac{\cot x}{1+\csc x}=\frac{\cos x(\sin x-1)}{-\cos^2 x}[/tex]
Simplify to get:
[tex]\frac{\cot x}{1+\csc x}=\frac{-(\sin x-1)}{\cos x}[/tex]
Or
[tex]\frac{\cot x}{1+\csc x}=\frac{1-\sin x}{\cos x}[/tex]
Divide both the numerator and denominator by sin x
[tex]\frac{\cot x}{1+\csc x}=\frac{\frac{1}{\sin x}-\frac{\sin x}{\sin x}}{\frac{\cos x}{\sin x}}[/tex]
This finally gives:
[tex]\frac{\cot x}{1+\csc x}=\frac{\csc x-1}{\cot x}[/tex]
