Trigonometric Functions — Maths STD 12 Science — Question

In ΔABC by sine rule, we have
$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}= k$
∴ a = k sin A, b = k sin B and c = k sin C
Now, consider
$\frac{a-b}{a+b}=\frac{k \sin A-k \sin B}{k \sin A+k \sin B}$
$=\frac{\sin A-\sin B}{\sin A+\sin B}$
$=\frac{2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)}{2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)}$
$=\cot \left(\frac{A+B}{2}\right) \cdot \tan \left(\frac{A-B}{2}\right)$
$=\cot \left(\frac{\pi}{2}-\frac{C}{2}\right) \cdot \tan \left(\frac{A-B}{2}\right) \ldots \ldots[\because A + B + C =\pi]$
$=\tan \left(\frac{C}{2}\right) \tan \left(\frac{A-B}{2}\right)$
$\therefore \frac{a-b}{a+b}=\tan \frac{C}{2} \cdot \tan \left(\frac{A-B}{2}\right)$
$\therefore \tan \left(\frac{A-B}{2}\right)=\frac{a-b}{a+b} \cot \left(\frac{C}{2}\right)$