Question

I get this far and get stuck on verifying the identity steps.

Answer 1

`csc(-x)-1~~ = ~~\cfrac{cot^2(x)}{csc(-x)+1} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{cot^2(x)}{csc(-x)+1}\implies \cfrac{~~ ( cos^2(x))/(sin^2(x) ) ~~}{(1)/(sin(-x))+1}\implies \cfrac{~~ ( cos^2(x))/(sin^2(x) ) ~~}{(1)/(-sin(x))+1}\implies \cfrac{~~ ( cos^2(x))/(sin^2(x) ) ~~}{1-(1)/(sin(x))} \\\\\\ \cfrac{~~ ( cos^2(x))/(sin^2(x) ) ~~}{(sin(x)-1)/(sin(x))}\implies \cfrac{ cos^2(x)}{sin^2(x) }\cdot \cfrac{sin(x)}{sin(x)-1}\implies \cfrac{ cos^2(x)}{sin(x)[sin(x)-1] }`

`\cfrac{ cos^2(x)}{-sin(x)[1-sin(x)] }\implies \cfrac{ -cos^2(x)}{sin(x)[1-sin(x)] } \\\\\\ \cfrac{ -cos^2(x)}{sin(x)[1-sin(x)] }\cdot \cfrac{1+sin(x)}{1+sin(x)}\implies \cfrac{-cos^2(x)[1+sin(x)]}{sin(x)\underset{ \textit{difference of squares} }{[1-sin(x)][1+sin(x)]}} \\\\\\ \cfrac{-cos^2(x)[1+sin(x)]}{sin(x)[1-sin^2(x)]}\implies \cfrac{-cos^2(x)[1+sin(x)]}{sin(x)[cos^2(x)]}`

`\cfrac{-[1+sin(x)]}{sin(x)}\implies \cfrac{-1-sin(x)}{sin(x)}\implies \cfrac{-1}{sin(x)}-\cfrac{sin(x)}{sin(x)}\implies \cfrac{1}{-sin(x)}-1 \\\\\\ \cfrac{1}{sin(-x)}-1\implies \boxed{csc(-x)-1 }`