Answer
$$(\sec\alpha+\csc\alpha)(\cos\alpha-\sin\alpha)=\cot\alpha-\tan\alpha$$
The expression is proved below to be an identity.
Work Step by Step
$$(\sec\alpha+\csc\alpha)(\cos\alpha-\sin\alpha)=\cot\alpha-\tan\alpha$$
We examine the left side.
$$A=(\sec\alpha+\csc\alpha)(\cos\alpha-\sin\alpha)$$
$\sec\alpha$ and $\csc\alpha$ can be rewritten as follows.
$$\sec\alpha=\frac{1}{\cos\alpha}\hspace{2cm}\csc\alpha=\frac{1}{\sin\alpha}$$
which means,
$$A=\Big(\frac{1}{\cos\alpha}+\frac{1}{\sin\alpha}\Big)(\cos\alpha-\sin\alpha)$$
$$A=\frac{\sin\alpha+\cos\alpha}{\sin\alpha\cos\alpha}(\cos\alpha-\sin\alpha)$$
$$A=\frac{(\cos\alpha+\sin\alpha)(\cos\alpha-\sin\alpha)}{\sin\alpha\cos\alpha}$$
$$A=\frac{\cos^2\alpha-\sin^2\alpha}{\sin\alpha\cos\alpha}$$ (as $(a+b)(a-b)=a^2-b^2$)
$$A=\frac{\cos^2\alpha}{\sin\alpha\cos\alpha}-\frac{\sin^2\alpha}{\sin\alpha\cos\alpha}$$
$$A=\frac{\cos\alpha}{\sin\alpha}-\frac{\sin\alpha}{\cos\alpha}$$
$$A=\cot\alpha-\tan\alpha\hspace{1cm}\text{(Quotient Identities)}$$
2 sides are equal, so the expression is an identity.
