Question

Prove cscØ/cotØ - cotØ/cscØ=sin^2Ø/cosØ

Answer 1

probably no trig at all involved with this, just algebra
if you put \(\cos(\theta)=a\) and \(\sin(\theta)=b\) then you are supposed to show that
\[\large \frac{\frac{1}{b}}{\frac{a}{b}}-\frac{\frac{a}{b}}{\frac{1}{b}}=\frac{b^2}{a}\]

Answer 2

can you do that algebra?

Answer 3

yes, but i;m having difficulty coming up with the same denoms

Answer 4

rewrite the left hand sides as \[\frac{b}{ab}-\frac{ab}{b}\]

Answer 5

common denominator is \(ab\) so you get
\[\frac{b}{ab}-\frac{a^2b}{ab}=\frac{b-a^2b}{ab}\]

Answer 6

factor and cancel, gets
\[\frac{b-a^2b}{ab}=\frac{b(1-a^2}{ab}=\frac{1-a^2}{a}\]

Answer 7

oh so there is one tiny piece of trig
now we have
\[\frac{1-\cos^2(x)}{\cos(x)}\] one tiny trig step and you are done