If $BC=a$, $AC=b$, and $AB=c$ are the sides of a triangle $ABC$, and $\angle C\ne \frac{\pi}{2}$, then which one of the following is not correct?

By sine rule,

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

So,

$\frac{a-b}{a+b}=\frac{\sin A-\sin B}{\sin A+\sin B}$

Now,

$\sin A-\sin B=2\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)$

and

$\sin A+\sin B=2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$

Therefore,

$\frac{\sin A-\sin B}{\sin A+\sin B}=\cot\left(\frac{A+B}{2}\right)\tan\left(\frac{A-B}{2}\right)$

Since,

$A+B=\pi-C$

So,

$\cot\left(\frac{A+B}{2}\right)=\cot\left(\frac{\pi-C}{2}\right)$

$=\cot\left(\frac{\pi}{2}-\frac{C}{2}\right)$

$=\tan\left(\frac{C}{2}\right)$

Thus,

$\frac{a-b}{a+b}=\tan\left(\frac{C}{2}\right)\tan\left(\frac{A-B}{2}\right)$

But option $4$ gives

$\frac{a-b}{a+b}=\frac{\tan\left(\frac{A-B}{2}\right)}{\tan\left(\frac{C}{2}\right)}$