Trigonometric proof

**dagbayani481** · March 22nd 2011, 11:20 PM

Hey guys, in need of help again. I feel seriously stumped and I don't even know if I'm doing this right but anyways. Here is the proof:

$secxcscx(tanx+cotx)=sec^2x+csc^2x$

Next I changed the $(secx*cscx)$
to $\frac{cscx}{cosx}$
After that I distributed and got
$\frac{cscxtanx}{cosx}+\frac{cotxcscx}{cosx}$

From then on I'm lost, and honestly I don't even know if I did the right thing here.
Help will be greatly appreciated!

**Prove It** · March 22nd 2011, 11:26 PM

$\displaystyle \sec{x}\csc{x}(\tan{x} + \cot{x}) \equiv \frac{1}{\cos{x}}\left(\frac{1}{\sin{x}}\right)\le ft(\frac{\sin{x}}{\cos{x}} + \frac{\cos{x}}{\sin{x}}\right)$

$\displaystyle \equiv \frac{1}{\sin{x}\cos{x}}\left(\frac{\sin^2{x}}{\si n{x}\cos{x}} + \frac{\cos^2{x}}{\sin{x}\cos{x}}\right)$

$\displaystyle \equiv \frac{1}{\sin{x}\cos{x}}\left(\frac{\sin^2{x} + \cos^2{x}}{\sin{x}\cos{x}}\right)$

$\displaystyle \equiv \frac{1}{\sin{x}\cos{x}}\left(\frac{1}{\sin{x}\cos {x}}\right)$

$\displaystyle \equiv \frac{1}{\sin^2{x}\cos^2{x}}$

$\displaystyle \sec^2{x}\csc^2{x}$ .

I disagree with your RHS...

**dagbayani481** · March 22nd 2011, 11:53 PM

Ah, I see! Thank you very much!
I enjoy doing identities (No lie) but my only gripe is trying to figure out where to start. I think it's the fact that there are so many possibilities. Are there any tips on figuring out the best method? My teacher told me to start with the side with the most terms, but other then that I usually have a hard time trying to decipher what to do next.

**Prove It** · March 22nd 2011, 11:54 PM

I usually convert things to sines and consines, as they're usually the easiest to work with...

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