Answer
$$\sec x-\cos x+\csc x-\sin x-\sin x\tan x=\cos x\cot x$$
The expression is indeed an identity.
Work Step by Step
$$\sec x-\cos x+\csc x-\sin x-\sin x\tan x=\cos x\cot x$$
We examine the left side first.
$$A=\sec x-\cos x+\csc x-\sin x-\sin x\tan x$$
What we have here is a mixture of many different types of trigonometric expressions. Since there are no clear clues as to what to do, it is probably best to rewrite all of them in terms of $\sin x$ and $\cos x$.
$$A=\frac{1}{\cos x}-\cos x+\frac{1}{\sin x}-\sin x-\sin x\frac{\sin x}{\cos x}$$
$$A=\frac{1-\cos^2 x}{\cos x}+\frac{1-\sin^2x}{\sin x}-\frac{\sin^2x}{\cos x}$$
$$A=\frac{(1-\cos^2x)-\sin^2x}{\cos x}+\frac{1-\sin^2x}{\sin x}$$
Notice here that we can rewrite $1-\cos^2x=\sin^2x$ and $1-\sin^2x=\cos^2x$
$$A=\frac{\sin^2x-\sin^2x}{\cos x}+\frac{\cos^2x}{\sin x}$$
$$A=\frac{0}{\cos x}+\frac{\cos^2x}{\sin x}$$
$$A=\frac{\cos^2x}{\sin x}$$
Now we try to make $A$ into $\cos x\cot x$.
$$A=\cos x\frac{\cos x}{\sin x}$$
$$A=\cos x\cot x$$ (as $\cot x=\frac{\cos x}{\sin x}$)
Therefore the given expression has been proved to be an identity.
