Question

Factor the algebraic expression below in terms of a single trigonometric function.

csc^2x - 1

Answer 1

I'm from Europe (Holland). Back here, we do not use sec ans csc functions, so I would use sin and cos:\[\csc ^2x-1=\frac{ 1 }{ \sin ^2x }-1=\frac{ \sin^2x+\cos ^2 x }{ \sin ^2x }-1=1+\frac{ \cos^2x }{ \sin^2x }-1\]Now you can see where this is leading to, I guess...

Answer 2

Not really, sorry, not good at all at this.

Answer 3

OK, if you look at the last step of my calculation, you'll see 1 + .... -1, so 1-1=0 and they are gone, leaving:\[\frac{ \cos^2x }{ \sin^2x }=\left( \frac{ \cos x }{ \sin x } \right)^2=\cot^2x\]and that is your single trigonometric function!

Answer 4

Oh, thanks!