Explanation

(b) $\overrightarrow{A} = A\widehat{A} = B\widehat{B}.$ let $\theta$ be the angle between $\overrightarrow{A}$ and $\overrightarrow{B}.$
As per question,
$\cos\alpha = \frac{\left( A\widehat{A} + B\widehat{B} \right).(A\widehat{B} + B\widehat{A})}{\left| A\widehat{A} + B\widehat{B} \right|\left| A\widehat{B} + B\widehat{A} \right|}$
or $\cos\alpha = \frac{2AB + (A^{2} + B^{2})cos\theta}{\left( \sqrt{A^{2} + B^{2} + 2AB}cos\theta \right)^{2}}$
or $2AB + \left( A^{2} + B^{2} \right)cos\theta = (A^{2} + B^{2})\cos\alpha + 2ABcos\theta cos\alpha$
or $2AB(1 - cos\alpha cos\theta)$
$= \left( A^{2} + B^{2} \right)(cos\alpha - cos\theta)$
or $\frac{2AB}{A^{2} + B^{2}} = \frac{cos\alpha - cos\theta}{1 - cos\alpha cos\theta}$
or $\frac{2AB}{(A^{2} + B^{2})} = \frac{cos\alpha - cos\theta}{1 - cos\alpha cos\theta}$
or $\frac{2AB + (A^{2} + B^{2})}{\left( A^{2} + B^{2} \right) - AB}$
$= \frac{(cos\alpha cos\theta) + (1 - cos\alpha cos\theta)}{(1 - cos\alpha cos\theta) + (cos\alpha cos\theta)}$
or $\frac{(A + B)^{2}}{(A - B)^{2}} = \frac{(1 + cos\alpha)(1 - cos\theta)}{(1 + cos\theta)(1 - cos\alpha)}$
$= \frac{\tan^{2}{\theta/2}}{\tan^{2}{\theta/2}}$
or tan$\frac{\alpha}{2} = \left( \frac{A - B}{A + B} \right)\tan\frac{\theta}{2}$