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