Question

Where would I start to prove cotx * cosx = cscx - sinx is an identity?

Answer 1

In what other ways can cot(x) and csc(x) be written?

Answer 2

cotx = cosx/sinx and cscx = 1/sinx

Answer 3

Substitute one of those into the original equation and start solving it.

Answer 4

That's where I am right now, but I'm wondering if there's an identity or someting I can use to help solve it. Right now I've simplified everything in terms of cos and sin values

Answer 5

\[cosx/sinx * cosx = 1/sinx - sinx\] that's how far I am

Answer 6

Is \[(cotx + 1)^{2} \] the same thing as \[\cot ^{2}x + 1\] ?

Answer 7

no

Answer 8

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

Since the pythagorean identity states:
\[(\sin x)^{2}+(\cos x)^{2}=1\]

Answer 9

Keep working it from there.