# Trigonometric proof

• Mar 22nd 2011, 11:20 PM
dagbayani481
Trigonometric proof
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!
• Mar 22nd 2011, 11:26 PM
Prove It
$\displaystyle \sec{x}\csc{x}(\tan{x} + \cot{x}) \equiv \frac{1}{\cos{x}}\left(\frac{1}{\sin{x}}\right)\le ft(\frac{\sin{x}}{\cos{x}} + \frac{\cos{x}}{\sin{x}}\right)$

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

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

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

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

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