# Trig Simplification 2

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• March 29th 2011, 05:55 PM
rtblue
Trig Simplification 2
Hey guys, I had another simplification question. You may start the process from either of these expressions, but the end result must the the other one:

$\displaystyle \frac{cscx}{secx-cscx}$

$\displaystyle \frac{\cot^2x+cscxsecx+1}{\tan^2x-\cot^2x}$

Any suggestions would be appreciated! Thanks in advance!
• March 29th 2011, 06:03 PM
Prove It
I would start by converting everything to sines and cosines...
• March 29th 2011, 07:35 PM
Soroban
Hello, rtblue!

Quote:

$\displaystyle \frac{\csc x}{\sec x-\csc x} \;=\;\frac{\cot^2\!x+\csc x\sec x+1}{\tan^2\!x-\cot^2\!x}$

I started with the right side:

$\displaystyle\frac{\cot^2\!x + \csc x\sec x + 1}{\tan^2\!x - \cot^2\!x} \;=\; \frac{\overbrace{\cot^2\!x + 1}^{\text{This is }\csc^2\!x} + \csc x\sec x}{(\sec^2\!x - 1) - (\csc^2\!x - 1)}$

. . $\displaystyle =\;\frac{\csc^2\!x + \csc x \sec x}{\sec^2\!x - 1 - \csc^2\!x + 1} \;=\;\frac{\csc x\sec x + \csc^2\!x}{\sec^2\!x - \csc^2\!x}$

. . $\displaystyle=\; \frac{\csc x(\sec x + \csc x)}{(\sec x - \csc x)(\sec x + \csc x)} \;=\;\frac{\csc x}{\sec x - \csc x}$