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:
$$\frac{\cos(x)}{\frac{1}{\cos(x)}\cdot\sin(x)}=\frac{\cos(x)}{\frac{\sin(x)}{\cos(x)}}=\frac{\cos(x)}{1}\cdot\frac{\cos(x)}{\sin(x)}=\frac{\cos^2(x)}{\sin(x)}$$
 A: Yes, you are on the right track. Use the relation $$\cos^2 x + \sin^2 x = 1$$ to achieve the result from here. Good work! Just beware you had a $\theta$ lying in your second line and you need to add explanations, otherwise as a draft it's OK.
A: Yes, you are on the right track. Now, you want to turn $\frac{\cos^2 x}{\sin x}$ into $\csc x - \sin x$. Do you know of any identities using $\cos^2 x$ that might help turn that into the difference of two things?
A: 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
$$\sin^2(x) + \cos^2(x) = 1$$

$$\begin{align*}
\frac{\cos x}{\sec x \sin x} & = \frac{\cos x}{\frac1{\cos x}\sin x} = \frac{\cos^2 x}{\sin x} \\
& = \frac{1-\sin^2 x}{\sin x} = \frac1{\sin x} - \sin x\\
& = \csc x - \sin x
\end{align*}$$
A: First, we'll expand $\sec{x} = \frac{1}{\cos{x}}$.
Let's begin!
$$\require{cancel}\begin{aligned}\frac{\cos{x}}{\sec{x}\sin{x}}&=\frac{\cos{x}}{\frac{1}{\cos{x}}\cdot\sin{x}}\\&=\frac{\cos{x}}{\frac{\sin{x}}{\cos{x}}}\\&=\cos{x}\cdot\frac{\cos{x}}{\sin{x}}\\&=\frac{\cos^2{x}}{\sin{x}}\\&=\frac{1-\sin^2{x}}{\sin{x}}\\&=\frac{1}{\sin{x}}-\frac{\cancelto{\sin{x}}{\sin^2{x}}}{\cancelto{1}{\sin{x}}}\\&=\csc{x}-\sin{x}\end{aligned}$$
I hope this helps.
A: Use $\cos ^2 x + \sin ^2x =1$, that should do the trick.
A: I think that many students don't understand that finding the answer is different from presenting the answer. I call this method the meet-in-the-middle method. Take the left-hand-side (LHS) and right-hand-side (RHS) and try to show that they are equal, but make sure that any operations you perform don't change the value of the expression (e.g. multiply by one or add zero). This would typically be done on scrap paper. Then to present the argument, copy the steps for the LHS argument forward and the steps for the RHS backwards.
For example, here's my scrap paper:
$$\frac{\cos x}{\sec x \sin x} = \csc x - \sin x$$
$$\frac{\cos x}{\frac{1}{\cos x} \sin x} = \frac{1}{\sin x} - \sin x$$
$$\frac{\cos^2 x}{\sin x} = \frac{1-\sin^2x}{\sin x}$$
$$\frac{\cos^2 x}{\sin x} = \frac{\cos^2 x}{\sin x}$$
Now to present the answer, copy the steps from the LHS forward, and the steps from the RHS backwards.
$$\frac{\cos x}{\sec x \sin x}=\frac{\cos x}{\frac{1}{\cos x} \sin x}=\frac{\cos^2 x}{\sin x}=\frac{1-\sin^2x}{\sin x}=\frac{1}{\sin x} - \sin x=\csc x - \sin x$$
This method almost always works, as long as you know your identities.
For an easy way to remember the Pythagorean identities, see my answer at Basic trigonometry identities question
