1. ## Simple trig

Can someone refresh my memory on how i transform:
from: $y\prime = \csc(x)\cdot\cot(x) - \cos(x)$
to: $y\prime = \cos(x)\cot^2(x)$

I'm thinking:
$y\prime = (\frac{\cos(x)}{\sin(x)}) - (\frac{\cos(x)}{1})$

then factor to get
$y\prime = \cos(x)(\frac{1}{\sin^2(x)}-1)$

then
$y\prime = \cos(x)(\csc^2(x)-1)$

but i'm not sure if thats right and if it is how do i get the $\cot^2(x)$ from the $(\csc^2(x)-1)$?

Thanks

2. Originally Posted by ixo
Can someone refresh my memory on how i transform:
from: $y\prime = \csc(x)\cdot\cot(x) - \cos(x)$
to: $y\prime = \cos(x)\cot^2(x)$

I'm thinking:
$y\prime = (\frac{\cos(x)}{\sin(x)}) - (\frac{\cos(x)}{1})$

then factor to get
$y\prime = \cos(x)(\frac{1}{\sin^2(x)}-1)$

but i'm not sure if thats right and if it is how do i get the $\cot^2(x)$ from the $(\frac{1}{\sin^2(x)}-1)$?

Thanks
Recall that $\sin^2 x + \cos^2 x = 1 \implies 1 + \frac {\cos^2 x}{\sin^2 x} = \frac 1{\sin^2 x}$ $\implies \frac {\cos^2 x}{\sin^2 x} = \frac 1{\sin^2 x} - 1 \implies \cot^2 x = \csc^2 x - 1$

3. $y\prime = \csc(x)\cdot\cot(x) - \cos(x) = \frac{1}{\sin x}\cdot\frac{\cos x}{\sin x} - \cos(x) = \frac{\cos x (1-\sin^2x)}{sin^2x} = \frac{\cos^3x}{\sin^2x}$

$y\prime = \cos(x)\cot^2(x)$

EDIT : too late !