# Roadblock in trig identity

1. Oct 13, 2006

### Checkfate

Hi,

I am trying to prove $$\frac{tanx+1}{tanx-1}=\frac{secx+cscx}{secx-cscx}$$

I am working on the right side and trying to get it to the same form as the left side.

I get to $$\frac{1+2sinxcosx}{sin^{2}x-cos^{2}x}$$ which I COULD apply the fact that sin2x=2sinxcosx as well as cos^2x-sin^2x=cos2x (after multiplying both top and bottom by -1) but that gets me to

$$\frac{-Sin(2x)-1}{cos(2x)}$$ which looks completely useless!

Anyone see something I don't? Just point me in the right direction :) Thanks again

2. Oct 13, 2006

### Checkfate

oh my god why didn't I see that! lol. I was looking at that step ( I had the exact same work shown) and thought that was an important step, but dividing by cos x just never popped into my head, lol. Anyways, thanks a ton :).

3. Oct 13, 2006

### HallsofIvy

Staff Emeritus
Actually, starting from
$$\frac{tanx+1}{tanx-1}=\frac{secx+cscx}{secx-cscx}$$
and multiplying the numerator and denominator of the right side by sin x gives the result immediately.

4. Oct 13, 2006

### Checkfate

lol, how do you see this stuff? I don't even start thinking about solving it until at least the 3rd step. Maybe I answered my own question here... lol :)

Thanks to both of you, I am going to start looking for a way to immediately put it into the desired form at EVERY step from now on.

Last edited: Oct 13, 2006

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