How do you prove:
csc^2@cos^2@ = csc^2@ -1
\(cscx=\frac{1}{sinx};\ sin^2x+cos^2x=1\to cos^2x=1-sin^2x\\----------------------------\\\\csc^2xcos^2x=csc^2x-1\\\\L=csc^2xcos^2x=\frac{1}{sin^2x}\cdot cos^2x=\frac{cos^2x}{sin^2x}\\\\R=csc^2x-1=\frac{1}{sin^2x}-1=\frac{1}{sin^2x}-\frac{sin^2}{sin^2x}=\frac{1-sin^2x}{sin^2x}=\frac{cos^2x}{sin^2x}\\\\\boxed{L=R}\)
\(LHS\\ \\ ={ cosec }^{ 2 }\theta { cos }^{ 2 }\theta \\ \\ ={ cosec }^{ 2 }\theta \left( 1-{ sin }^{ 2 }\theta \right)\)
\(\\ \\ ={ cosec }^{ 2 }\theta -{ cosec }^{ 2 }\theta { sin }^{ 2 }\theta \\ \\ ={ cosec }^{ 2 }\theta -\frac { 1 }{ { sin }^{ 2 }\theta } \cdot { sin }^{ 2 }\theta\)
\(\\ \\ ={ cosec }^{ 2 }\theta -1\\ \\ =RHS\)