I have done most of these problems...If someone could please look at my work, I would really appreciate it...There were several that I have had trouble with as well that are incomplete...please help...
Hello, aikenfan!
The ones you completed are all correct . . . Nice work!
$\displaystyle 17)\;\;\frac{1}{\sec x\tan x} \:=\:\csc x - \sin x$
The left side is: .$\displaystyle \frac{1}{\dfrac{1}{\cos x}\cdot\dfrac{\sin x}{\cos x}} \;=\;\frac{\cos^2\!x}{\sin x} \;=\;\frac{1-\sin^2\!x}{\sin x} \;= \;\frac{1}{\sin x} - \frac{\sin^2\!x}{\sin x} \;=\;\csc x - \sin x$
$\displaystyle 20)\;\;\sec x - \cos x \:=\:\sin x\tan x$
The left side is: .$\displaystyle \frac{1}{\cos x} - \cos x \;=\;\frac{1-\cos^2\!x}{\cos x} \;=\;\frac{\sin^2\!x}{\cos x} \;=\;\sin x\cdot\frac{\sin x}{\cos x} \;=\;\sin x\tan x$