# Thread: Another Trig Identity problem.

1. ## Another Trig Identity problem.

I need to know in steps how to prove this identity, if anyone minds helping. Thanks!

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

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

3. Originally Posted by harish21
$\displaystyle \tan(x)=\dfrac{\sin(x)}{\cos(x)}$

$\displaystyle \therefore \cos(x)\sin(x)\tan(x)=\cos(x)\sin(x)\dfrac{\sin(x) }{\cos(x)}=....$
Can't solve?

4. Are you even attempting?

You should be able to recognize that the cos(x) in the numerator and the denominator cancel each other out. So you are left with $\displaystyle \cos(x)\sin(x)\dfrac{\sin(x)}{\cos(x)} = \sin(x)\sin(x)=\sin^2(x)$

You have been told in your previous post that $\displaystyle sin^2(x)+cos^2(x)=1$. based on that, $\displaystyle \sin^2(x)= ??$

5. I didn't realize they canceled each other out, but when I did the equation myself I canceled them out anyway. When I was left with $sin^2x&space;=&space;1-cos^2x$ I just added the negative to the other side to make a positive, thus making it all equal one. I thought I was doing it wrong though, since I didn't know they canceled. Thanks for clarification.