Help on this TRIG IDENTITY problem (yes i know its not SAT)

<p>Math is so confusing. I hate it. :(</p>

<p>Someone put up another problem. It's just a good preparation here. Actually I forgot the universal substitution. kk.
Btw, do I have to get TI89 (is name correct??) or sth like that to take SATII. is calculator must-have?? I have a very simple calculator. So can I manage with it?</p>

<p>One of the most important aspects of proof is elegance. Having said that, consider this approach</p>

<p>Prove: </p>

<p>(1 + cosx + sinx)/(1 + cosx - sinx) = tanx + secx -----(1)</p>

<p>Sol:</p>

<p>Let 1+cosx = u, sinx = v ------------(2)</p>

<p>Substituting eq. (2) into LHS of eq. (1), we get</p>

<p>u + v/ u-v
= (u^2+2uv +v^2)/(u^2 –v^2) ----------(3)</p>

<p>Substituting eq. (2) into eq. (3), we get</p>

<p>(1+2cosx+cos^2 x+2cosxsinx + 2 sinx +sin^2x)/(1+2cosx +cos^2x – sin^2x)</p>

<p>=(cosx+1)(sinx+1)/cosx(cosx+1)</p>

<p>= tanx + secx ---------------------(4)</p>

<p>Eq.(4) is indeed the same as the RHS of eq.(1) ==> Q.E.D.</p>

<p>Wow, it's the easiest approach!!!</p>

<p>
[Quote]

Actually, you can start with something and show that it leads to something true, and that will prove the original statement. There is fine print though:
all the transformations must be identical.

[/Quote]
</p>

<p>Well, by identical I'm assuming you mean reversible and one to one. And generally, when you do that, the proof itself is the series of steps, but in reverse. That is to say, once you go from what you're trying to prove to an identity, the actual proof is everything you've done backwards.</p>

<p>^^Exactly.</p>