Precalc - Prove Trig Identity I'm trying to prove the following problem:
$$\dfrac{\sec^2(x)}{\cot(x)} - \tan^3(x) = \tan(x)$$
I thought I had the answer, but the internet says otherwise. Unfortunately I have been unable to find step-by-step problems to show where I went wrong.
 A: Put it all in Tan functions.
$$\sec^2x+1 = tan^2x$$
$$\frac{1}{cotx} =tanx$$
A: We will work on the left side and show it is equl to $\tan(x)$. Let's change everything into $\sin(x)$ and $\cos(x)$ using the following formulas
$$\sec^2(x) = \frac{1}{\cos^2(x)}, \quad \quad \cot(x) = \frac{\cos(x)}{\sin(x)}, \quad \quad \tan^3(x) = \frac{\sin^3(x)}{\cos^3(x)}.$$
This will give us:
$$\frac{\frac{1}{\cos^2(x)}}{\frac{\cos(x)}{\sin(x)}} - \frac{\sin^3(x)}{\cos^3(x)}.$$
Cleaning up the left fraction will give you 
$$\frac{\sin(x)}{\cos^3(x)} -  \frac{\sin^3(x)}{\cos^3(x)}.$$
From here, factor out $\sin(x)$ from the top to get:
$$\frac{\sin(x)(1 - \sin^2(x))}{\cos^3(x)}.$$
Finally, use the Pythagorean Identity $\cos^2(x) = 1 - \sin^2(x)$ to finish it off.
A: Hint: $\sec^2 x + 1 = \tan^2 x$. Start with the left side and simplify using this identity.
If you'd like feedback regarding where you went wrong (or for us to check your work), feel free to edit your post to include your solution!
A: Hint: $\displaystyle \frac{1}{\cot x} = \tan x$. So we get $\tan x \sec^2 x - \tan^3 x = \tan x$. Divide the equation by a certain trigonometric function and you should get a familiar equation.
