Step 1: Set up the goal.
We must find which identity for a triangle is NOT correct. We will build the genuine relation for $\frac{a-b}{a+b}$ and then see which option disagrees with it.
Step 2: Convert sides to sines.
By the Law of Sines, $a\propto\sin A$ and $b\propto\sin B$, so $\dfrac{a-b}{a+b}=\dfrac{\sin A-\sin B}{\sin A+\sin B}$. This already validates option (C).
Step 3: Apply sum-to-product formulas.
Using $\sin A-\sin B=2\cos\frac{A+B}{2}\sin\frac{A-B}{2}$ and $\sin A+\sin B=2\sin\frac{A+B}{2}\cos\frac{A-B}{2}$, the ratio becomes $\cot\frac{A+B}{2}\tan\frac{A-B}{2}$, which validates option (A).
Step 4: Bring in the angle $C$.
Since $A+B=\pi-C$, we have $\frac{A+B}{2}=\frac{\pi}{2}-\frac{C}{2}$, so $\cot\frac{A+B}{2}=\tan\frac{C}{2}$. Hence the true relation is $\dfrac{a-b}{a+b}=\tan\frac{C}{2}\tan\frac{A-B}{2}$.
Step 5: Test option (B).
Option (B) claims $\dfrac{a-b}{a+b}=\dfrac{\tan\frac{A-B}{2}}{\tan\frac{C}{2}}$, which divides by $\tan\frac{C}{2}$ instead of multiplying. This contradicts the correct relation found in Step 4.
Step 6: Confirm option (D) is fine.
Option (D) is just the Law of Sines, which is always true. So the only wrong statement is (B).
\[ \boxed{\dfrac{a-b}{a+b}=\dfrac{\tan\left(\dfrac{A-B}{2}\right)}{\tan\left(\dfrac{C}{2}\right)}} \]