Question

How do I factor the algebraic expression, csc^2(x) - 1, in terms of a single trigonometric function?

Answer 1

`\csc^2(x)-1=\left((1)/(\sin(x))\right)^2-1=(1)/(\sin^2(x))-1\\\\=(1)/(\sin^2(x))-(\sin^2(x))/(\sin^2(x))=(1-\sin^2(x))/(\sin^2(x))=(\cos^2(x))/(\sin^2(x))=\left((\cos(x))/(\sin(x))\right)^2\\\\=\cot^2(x)\\\\\boxed{\csc^2(x)-1=\cot^2(x)}\\\\Used:\\\\\csc(x)=(1)/(\sin(x))\\\\\sin^2(x)+\cos^2(x)=1\to\cos^2(x)=1-\sin^2(x)\\\\(\cos(x))/(\sin(x))=\cot(x)`