If $\;$ $\tan \phi = \left(\dfrac{a - b}{a + b}\right) \cot \left(\dfrac{C}{2}\right)$, $\;$ prove that $\;$ $c = \left(a + b\right) \left[\dfrac{\sin \left(\dfrac{C}{2}\right)}{\cos \phi}\right]$
Given: $\;$ $\tan \phi = \left(\dfrac{a - b}{a + b}\right) \cot \left(\dfrac{C}{2}\right)$ $\;\;\; \cdots \; (1)$
By cosine rule, $\;$ $\cos C = \dfrac{a^2 + b^2 - c^2}{2 a b}$
$\begin{aligned}
i.e. \; c^2 & = a^2 + b^2 - 2 a b \cos C \\\\
& \left[\text{Note: }\sin^2 \left(\dfrac{\theta}{2}\right) + \cos^2 \left(\dfrac{\theta}{2}\right) = 1 \right. \\
& \left. \hspace{1.2cm} \cos \theta = \cos^2 \left(\dfrac{\theta}{2}\right) - \sin^2 \left(\dfrac{\theta}{2}\right) \right] \\\\
& = \left(a^2 + b^2\right) \left[\sin^2 \left(\dfrac{C}{2}\right) + \cos^2 \left(\dfrac{C}{2}\right)\right] - 2 a b \left[\cos^2 \left(\dfrac{C}{2}\right) - \sin^2 \left(\dfrac{C}{2}\right)\right] \\\\
& = \sin^2 \left(\dfrac{C}{2}\right) \left[a^2 + b^2 + 2 a b\right] + \cos^2 \left(\dfrac{C}{2}\right) \left[a^2 + b^2 - 2 a b\right] \\\\
& = \left(a + b\right)^2 \sin^2 \left(\dfrac{C}{2}\right) + \left(a - b\right)^2 \cos^2 \left(\dfrac{C}{2}\right) \\\\
& = \left(a + b\right)^2 \sin^2 \left(\dfrac{C}{2}\right) \left[1 + \left(\dfrac{a - b}{a + b}\right)^2 \times \dfrac{\cos^2 \left(\dfrac{C}{2}\right)}{\sin^2 \left(\dfrac{C}{2}\right)}\right] \\\\
& = \left(a + b\right)^2 \sin^2 \left(\dfrac{C}{2}\right) \left[1 + \left(\dfrac{a - b}{a + b}\right)^2 \cot^2 \left(\dfrac{C}{2}\right)\right] \\\\
& = \left(a + b\right)^2 \sin^2 \left(\dfrac{C}{2}\right) \left[1 + \tan^2 \phi\right] \;\;\; \left[\text{by equation (1)}\right] \\\\
& = \left(a + b\right)^2 \sin^2 \left(\dfrac{C}{2}\right) \sec^2 \phi \\\\
\therefore \; c & = \left(a + b\right) \sin \left(\dfrac{C}{2}\right) \sec \phi \\\\
i.e. \; c & = \left(a + b\right) \left[\dfrac{\sin \left(\dfrac{C}{2}\right)}{\cos \phi}\right]
\end{aligned}$
Hence proved.