I'm trying to verify this identity
$\sin^2 x \tan x = \tan x - \cos x \sin x$
Here is the correct answer
$\sin^2 x \tan x = (\sin^3/\cos x) = (\sin x/\cos x) - \cos x \sin x = \tan x - \cos x \sin x$
What is the logic behind:
$(\sin^3/\cos x) = (\sin x/\cos x) - \cos x \sin x $
How does $\cos x \sin x $ end up being subtracted from $(\sin x/\cos x)$?
I think what the answer you're looking at has failed at is being transparent. Recall the Pythagorean identity for trigonometry: $\sin^2\theta+\cos^2\theta=1$. This gives the equivalent form $\sin^2\theta=1-\cos^2\theta$ which is obtained just by subtracting $\cos^2\theta$ from both sides. Now... \begin{align*}(\sin^2\theta)(\tan\theta)&=(1-\cos^2\theta)(\tan\theta)\\ &=\tan\theta-\cos^2\theta\tan\theta\\ &=\tan\theta-\cos^2\theta\cdot\frac{\sin\theta}{\cos\theta}\\ &=\tan\theta-\cos\theta\sin\theta.\end{align*}
In the answer that you're looking at, they're using the fact that $\sin^3\theta=(\sin^2\theta)(\sin\theta)=(1-\cos^2\theta)(\sin\theta)$. At this point, you can distribute the $\sin\theta$ and split up the fraction based on the subtraction in the numerator to follow their work.