Overcoming Roadblocks in Trigonometric Identity Proofs

[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

 

[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 :).

 

[HallsofIvy]

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.

 

[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.