1. ## trig identity help...

I can't solve this :S

cosxsinx/tanx = 1 - sin^2x

thanks

2. Originally Posted by katiethegreat
I can't solve this :S

cosxsinx/tanx = 1 - sin^2x

thanks
Hint:

$\displaystyle \frac{\cos(x)\sin(x)}{\tan(x)}=\frac{\cos(x)\sin(x )}{1}\cdot \frac{\cos(x)}{\sin(x)}=...$

3. I don't know why....but i still don't get it :/

4. Originally Posted by katiethegreat
I don't know why....but i still don't get it :/
what is "it" you don't understand how I change what you had into this

$\displaystyle \frac{\cos(x)\sin(x)}{\tan(x)}=\frac{\cos(x)\sin(x )}{1}\cdot \frac{\cos(x)}{\sin(x)}=...$

or you don't know how to finish the problem. Where are you stuck?

5. both :/

6. Hi katiethegreat,

Maybe if you see it like this, it would help:

$\displaystyle \dfrac{\cos (x) \sin(x)}{\tan (x)}=$

$\displaystyle \dfrac{\cos(x) \sin(x)}{\frac{\sin (x)}{\cos (x)}}=$

$\displaystyle \dfrac{\cos (x) \sin (x)}{1} \cdot \frac{\cos (x)}{\sin (x)}=$

He just used the identity: $\displaystyle \tan (x) = \frac{\sin (x)}{\cos (x)}$

You should be able to finish now if you know $\displaystyle \sin^2(x)+\cos^2(x)=1$