Explanation

(b)
Since$,\ (a - b)\sin{(\theta + \phi) = (a + b)\sin{(\theta - \phi)}}$
$\Rightarrow a\left\{ \sin{(\theta + \phi) - \sin{(\theta - \phi)}} \right\} = b\left\{ \sin{(\theta - \phi) + \sin{(\theta + \phi)}} \right\}$
$\Rightarrow 2a\sin{\phi\cos{\theta = 2b\sin{\theta\cos\phi}}}$
$\Rightarrow a\tan{\phi = b\tan\theta}$
$\Rightarrow \frac{2a\tan\frac{\phi}{2}}{1 - \tan^{2}\frac{\phi}{2}} = \frac{2b\tan\frac{\theta}{2}}{1 - \tan^{2}\frac{\theta}{2}}$ …(i)
Since$,\ a\tan{\frac{\theta}{2} - b\tan{\frac{\phi}{2} = c}}$ (given)
$\Rightarrow \tan{\frac{\theta}{2} = \frac{b\tan{\frac{\phi}{2} + c}}{a}}$ …(ii)
From Eqs. (i) and (ii), we get
$\frac{a\tan\frac{\phi}{2}}{1 - \tan^{2}\frac{\phi}{2}} = \frac{b\frac{\left( b\tan{\frac{\phi}{2} + c} \right)}{a}}{1 - \frac{\left( b\tan{\frac{\phi}{2} + c} \right)^{2}}{a^{2}}}$
$\Rightarrow \tan{\frac{\phi}{2}(a^{2} - b^{2} - c^{2})} = bc\left( 1 + \tan^{2}\frac{\phi}{2} \right)$
Now$,\sin{\phi = \frac{2\tan\frac{\phi}{2}}{1 + \tan^{2}\frac{\phi}{2}} = \frac{2bc}{a^{2} - b^{2} - c^{2}}}$