Here we have the following expression:
[tex](csc x - 1) (csc x + 1)[/tex]
So we need to extend this. Remember that the difference of squares tells us:
[tex](a-b)(a+b)=a^2-b^2[/tex]
So here:
[tex]a=cscx \\ \\ b=1[/tex]
Thus:
[tex](csc x - 1) (csc x + 1)=csc^2x-1 \\ \\ \\ But: \\ \\ cscx=\frac{1}{sinx} \\ \\ \\ Then: \\ \\ csc^2x-1=\frac{1}{sin^2x}-1=\frac{1-sin^2x}{sin^2x}=\frac{cos^2x}{sin^2x}=cot^2x \\ \\ \\ Finally: \\ \\ \boxed{(csc x - 1) (csc x + 1)=cot^2x}[/tex]