In $\triangle ABC$, $\;$ prove that $\;$ $\dfrac{a + b}{a - b} = \tan \left(\dfrac{A + B}{2}\right) \cot \left(\dfrac{A - B}{2}\right)$
In $\;$ $\triangle ABC$, $\;$ $A + B + C = \pi$
i.e. $\;$ $A + B = \pi - C$
$\therefore \;$ $\dfrac{A + B}{2} = \dfrac{\pi}{2} - \dfrac{C}{2}$
$\therefore \;$ $\tan \left(\dfrac{A + B}{2}\right) = \tan \left(\dfrac{\pi}{2} - \dfrac{C}{2}\right) = \cot \left(\dfrac{C}{2}\right)$ $\;\;\; \cdots \; (1)$
$\begin{aligned}
\therefore \; RHS & = \tan \left(\dfrac{A + B}{2}\right) \cot \left(\dfrac{A - B}{2}\right) \\\\
& = \cot \left(\dfrac{C}{2}\right) \cot \left(\dfrac{A - B}{2}\right) \;\;\; \left[\text{by equation } (1)\right] \\\\
& = \dfrac{\cos \left(\dfrac{C}{2}\right) \cos \left(\dfrac{A - B}{2}\right)}{\sin \left(\dfrac{C}{2}\right) \sin \left(\dfrac{A - B}{2}\right)} \\\\
& = \dfrac{\dfrac{1}{2} \left[\cos \left(\dfrac{C + A - B}{2}\right) + \cos \left(\dfrac{C - A + B}{2}\right)\right]}{\dfrac{1}{2} \left[\cos \left(\dfrac{C - A + B}{2}\right) - \cos \left(\dfrac{C + A - B}{2}\right)\right]} \\\\
& = \dfrac{\cos \left(\dfrac{C + A}{2} - \dfrac{B}{2}\right) + \cos \left(\dfrac{C + B}{2} - \dfrac{A}{2}\right)}{\cos \left(\dfrac{C + B}{2} - \dfrac{A}{2}\right) - \cos \left(\dfrac{C + A}{2} - \dfrac{B}{2}\right)} \;\;\; \cdots \; (2)
\end{aligned}$
Now, in $\triangle ABC$,
$C + A = \pi - B$ $\implies$ $\dfrac{C + A}{2} = \dfrac{\pi}{2} - \dfrac{B}{2}$ $\;\;\; \cdots \; (3a)$
$C + B = \pi - A$ $\implies$ $\dfrac{C + B}{2} = \dfrac{\pi}{2} - \dfrac{A}{2}$ $\;\;\; \cdots \; (3b)$
$\therefore \;$ In view of equations $(3a)$ and $(3b)$, equation $(2)$ becomes
$\begin{aligned}
RHS & = \dfrac{\cos \left(\dfrac{\pi}{2} - \dfrac{B}{2} - \dfrac{B}{2}\right) + \cos \left(\dfrac{\pi}{2} - \dfrac{A}{2} - \dfrac{A}{2}\right)}{\cos \left(\dfrac{\pi}{2} - \dfrac{A}{2} - \dfrac{A}{2}\right) - \cos \left(\dfrac{\pi}{2} - \dfrac{B}{2} - \dfrac{B}{2}\right)} \\\\
& = \dfrac{\cos \left(\dfrac{\pi}{2} - B\right) + \cos \left(\dfrac{\pi}{2} - A\right)}{\cos \left(\dfrac{\pi}{2} - A\right) - \cos \left(\dfrac{\pi}{2} - B\right)} \\\\
& = \dfrac{\sin B + \sin A}{\sin A - \sin B} \;\;\; \cdots \; (4)
\end{aligned}$
By sine rule, $\;$ $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R$ $\;$ where $R$ is the circumradius of $\triangle ABC$
$\therefore \;$ $\sin A = \dfrac{a}{2 R}$ $\;\;\; \cdots \; (5a)$
and $\;$ $\sin B = \dfrac{b}{2 R}$ $\;\;\; \cdots \; (5b)$
$\therefore \;$ In view of equations $(5a)$ and $(5b)$, equation $(4)$ becomes
$RHS = \dfrac{\dfrac{b}{2R} + \dfrac{a}{2R}}{\dfrac{a}{2R} - \dfrac{b}{2R}} = \dfrac{a + b}{a - b} = LHS $
Hence proved.