1. ## Function Simplification

Hey guys, this problem has been racking my brain for around an hour. I guess I'm just having a brain fart, as it should be easy. Either way, here it is:

Simplify $\displaysyle \frac{cot^2x}{cscx+1}+1$

This should simplify out to csc(x). Could anyone explain how that happens?

2. never mind guys... I found it.

For those who want to know, just put everything under a common denominator. Then convert cot^2(x) + 1 to csc^2(x). Then factor, and it will cancel.

Thanks guys!

3. Originally Posted by rtblue
Hey guys, this problem has been racking my brain for around an hour. I guess I'm just having a brain fart, as it should be easy. Either way, here it is:

Simplify $\displaysyle \frac{cot^2x}{cscx+1}+1$

This should simplify out to csc(x). Could anyone explain how that happens?

I always start by replacing everything with sines and cosines:
$\displaystyle \frac{\frac{cos^2(x)}{sin^2(x)}}{\frac{1}{sin(x)} + 1} + 1$

$\displaystyle \frac{\frac{cos^2(x)}{sin^2(x)}}{\frac{1}{sin(x)} + 1} \cdot \frac{sin^2(x)}{sin^2(x)} + 1$

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

Now add the 1 to the fraction and simplify.

-Dan

4. Hello, rtblue!

It helps if you know this identity and its variations:

. . $\csc^2\theta \:=\:\cot^2\!\theta + 1$

$\displaystyle \text{Simplify: }\;\frac{\cot^2\!x}{\csc x+1}+1$

$\displaystyle \text{We have: }\:\frac{\cot^2\!x}{\csc x + 1} + 1$

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

. . . . . . $\displaystyle =\;\frac{(\csc x - 1)(\csc x + 1)}{\csc x + 1} + 1$

. . . . . . $=\;(\csc x - 1) + 1$

. . . . . . $=\; \csc x$