Trigonometric Expression Simplification Could someone explain how to simplify $(\cos(x)-\csc(x))/(\sin(x)-\sec(x))$? Any help would be appreciated.
 A: $$
\underbrace{\frac{\cos x - \dfrac 1 {\sin x}}{\sin x - \dfrac 1 {\cos x}} = \frac{(\cos x) (\sin x \cos x - 1)}{(\sin x)(\sin x \cos x - 1)}}_{\begin{smallmatrix} \text{Multiply both the numerator} \\
\text{and the denominator by $\sin x \cos x$.} \end{smallmatrix}} = \frac {\cos x}{\sin x} = \cdots
$$
A: You always want to start by writing everything out in terms of just sines and cosines:
$$
\frac {\cos x - \frac{1}{\sin x}}{\sin x - \frac{1}{\cos x}}
$$
It then usually helps to make this into a simpler fraction:
$$
\frac {\cos x - \frac{1}{\sin x}}{\sin x - \frac{1}{\cos x}} 
\frac{\cos x \sin x}{\cos x \sin x}= \frac{\cos^2 x \sin x - \cos x}{\sin^2 x \cos x - \sin x}
$$
At this point you usually need to use the Pythagorean relation $\sin^2 + \cos^2 = 1$, but this problem is easier, you just need to factor out $(\cos x \sin x -1)$ from the numerator and denominator, leaving 
$$
\frac{ \cos x }{\sin x} = \cot{x} 
$$
and that is the answer you want.
A: Hint: $\csc(x) = \dfrac{1}{\sin x}$, and $\sec(x) = \dfrac{1}{\cos x}$
A: ...and 
$$ 
\cos(x) - \frac{1}{\sin(x)} = \frac{\cos(x) \sin(x)-1}{\sin(x)}.
$$
Do something similar with the denominator, invert and multiply, and you should be on your way. 
