Answer
$$\frac{\cot s-\tan s}{\cos s+\sin s}=\frac{\cos s-\sin s}{\sin s\cos s}$$
The verification process is explained in detail below.
Work Step by Step
$$\frac{\cot s-\tan s}{\cos s+\sin s}=\frac{\cos s-\sin s}{\sin s\cos s}$$
The left side is examined first.
$$X=\frac{\cot s-\tan s}{\cos s+\sin s}$$
We can rewrite $\cot s$ and $\tan s$ according to the following identities:
$$\cot s=\frac{\cos s}{\sin s}\hspace{2cm}\tan s=\frac{\sin s}{\cos s}$$
$$X=\frac{\frac{\cos s}{\sin s}-\frac{\sin s}{\cos s}}{\cos s+\sin s}$$
$$X=\frac{\frac{\cos^2s-\sin^2s}{\sin s\cos s}}{\cos s+\sin s}$$
$$X=\frac{\cos^2s-\sin^2s}{\sin s\cos s(\cos s+\sin s)}$$
Now remember that $A^2-B^2=(A-B)(A+B)$. So if $A=\cos s$ and $B=\sin s$ like in $X$, we have
$$X=\frac{(\cos s-\sin s)(\cos s+\sin s)}{\sin s\cos s(\cos s+\sin s)}$$
$$X=\frac{\cos s-\sin s}{\sin s\cos s}$$
which is exactly the right side.
Therefore, the left side is equal to the right one. The equation $$\frac{\cot s-\tan s}{\cos s+\sin s}=\frac{\cos s-\sin s}{\sin s\cos s}$$
is an identity.
