First, remember the following equivalences:
[tex]cscx=\frac{1}{sinx}[/tex]and:
[tex]cos²x+sin²x=1[/tex]then, given the trigonometric equation, if we use the distributive property and the first equivalence on the left side, we get:
[tex]\begin{gathered} (csc²x-1)sin²x=(\frac{1}{sin²x}-1)sin²x \\ =\frac{sin²x}{sin²x}-sin²x=1-sin²x \end{gathered}[/tex]then, by the second equivalence:
[tex]\begin{gathered} 1-sin²x=cos²x \\ \Rightarrow cos²x=cos²x \end{gathered}[/tex]therefore, (csc²x-1)=cos²x
