## Trigonometry (11th Edition) Clone

$$\frac{\cot s-\tan s}{\cos s+\sin s}=\frac{\cos s-\sin s}{\sin s\cos s}$$ The verification process is explained in detail below.
$$\frac{\cot s-\tan s}{\cos s+\sin s}=\frac{\cos s-\sin s}{\sin s\cos s}$$ 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.
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