Question

csc^2(x) - 1

Answer 1

factor in terms of a single trigonometric function

Answer 2

\[\csc^2(x)-1=\frac{1}{\sin^2(x)}-1=\frac{1-\sin^2(x)}{\sin^2(x)}=\frac{\cos^2(x)}{\sin^2(x)}=\cot^2(x)\]

Answer 3

You sir, are god-sent. Could you explain where you get the 1 - sin^2(x) / sin^2(x)?

Answer 4

Haha thank man!
Just using the gifts God gave me!

when you have something in the form of:
\[\frac{a}{b}+\frac{c}{d}\]
You can put them under a common factor so like in the instance of:
\[\frac{1}{2}+\frac{3}{4}=\frac{2}{4}+\frac{3}{4}=\frac{2+3}{4}=\frac{5}{4}\]
In general, for any denominators \(b\) and \(d\), the best way to find the common denominator is usually \(bd\)
So we know that:
\[\frac{a}{b}+\frac{c}{d}=\left(\frac{a}{b}\times\frac{d}{d}\right)+\left(\frac{c}{d}\times\frac{b}{b}\right)=\frac{ad}{bd}+\frac{cb}{db}=\frac{ad+cb}{bd}\]
So recognize this scenario:
\[\frac{1}{\sin^2(x)}-1=\frac{1}{\sin^2(x)}+\frac{-1}{1}=\frac{1(1)+\sin^2(x)(-1)}{\sin^2(x)(1)}=\frac{1-\sin^2(x)}{\sin^2(x)}\]

Answer 5

That makes sense! Thanks a bunch! :)