<s>[INST] Verify algebraically that the equation $\frac{\cos(x)}{\sec(x)\sin(x)}=\csc(x)-\sin(x)$ is an identity I am stuck when I get to this point $\frac{\cos^2(x)}{\sin(x)}$. 
Am I on the right track? 

Verify algebraically that the equation is an identity:
$$\frac{\cos(x)}{\sec(x)\sin(x)}=\csc(x)-\sin(x)$$

My work:
$$\frac{\cos(x)}{\frac{1}{\cos(x)}\cdot\sin(x)}=\frac{\cos(x)}{\frac{\sin(x)}{\cos(x)}}=\frac{\cos(x)}{1}\cdot\frac{\cos(x)}{\sin(x)}=\frac{\cos^2(x)}{\sin(x)}$$
 [/INST] General hints:
Agree on only using $\sin$ and $\cos$ by substituting
$$\csc(x) = \frac1{\sin(x)}, \qquad \sec(x) = \frac1{\cos(x)}$$
Then get rid of any fractions by multiplying with the denominator and finally use
$$\sin^2(x) + \cos^2(x) = 1$$

$$\begin{align*}
\frac{\cos x}{\sec x \sin x} & = \frac{\cos x}{\frac1{\cos x}\sin x} = \frac{\cos^2 x}{\sin x} \\
& = \frac{1-\sin^2 x}{\sin x} = \frac1{\sin x} - \sin x\\
& = \csc x - \sin x
\end{align*}$$
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