Question

Do these equal

Answer 1

\[\frac{ 1-sinx }{ cosx } = \frac{ cosx }{ 1-sinx }\]

Answer 2

Do we have to use a different kind of method for this or what? :S

Answer 3

Are you sure you got the signs right? The second is just the inverse of the first, so it's essentially asking if:

\[x = \frac{1}{x}\]
This is only try if x = +/- 1, so the question then becomes:

\[\frac{1 - sinx}{cosx} = 1\]
Or:

\[1 - sinx = cosx\]\[1 = cosx + sinx\]
This can't be true, since cos^2 + sin^2 = 1. If you need to prove that, square both sides:

\[1 = (cosx + sinx)^{2}\]\[1 = cosx^{2} + 2cosxsinx + sinx^{2}\]\[1 = (cosx^{2} + sinx^{2}) + 2cosxsinx\]\[1 = 1 + 2cosxsinx\]\[0 = 2cosxsinx\]\[0 = sinxcosx\]
Now plug in the definition for sine and cosine:

\[0 = \frac{opp}{hyp}\frac{adj}{hyp}\]\[0 = \frac{(opp)(adj)}{hyp^{2}}\]\[0 = (opp)(adj)\]
This obviously isn't true for all cases so the answer is false.

There's probably an easier way to prove this!