and $\cos \theta + \cos \phi = b$…..$(ii)$
Squaring, ${\sin ^2}\theta + {\sin ^2}\phi + 2\sin \theta \sin \phi = {a^2}$
and ${\cos ^2}\theta + {\cos ^2}\phi + 2\cos \theta \cos \phi = {b^2}$
Adding, $2+ 2 $$(\sin \theta \sin \phi + \cos \theta \cos \phi ) = {a^2} + {b^2}$
==>$2\cos (\theta - \phi ) = {a^2} + {b^2} - 2$
==> $\cos (\theta - \phi ) = \frac{{{a^2} + {b^2} - 2}}{2}$
$ \Rightarrow \frac{{1 - {{\tan }^2}\frac{{\theta - \phi }}{2}}}{{1 + {{\tan }^2}\frac{{\theta - \varphi }}{2}}} = \frac{{{a^2} + {b^2} - 2}}{2}$
==> $({a^2} + {b^2}) + ({a^2} + {b^2}){\tan ^2}\frac{{\theta - \phi }}{2} - 2 - 2{\tan ^2}\frac{{\theta - \phi }}{2}$
$ = 2 - 2{\tan ^2}\frac{{\theta - \varphi }}{2}$
==>$\frac{{4 - {a^2} - {b^2}}}{{{a^2} + {b^2}}} = {\tan ^2}\frac{{\theta - \phi }}{2}$
==> $\tan \frac{{(\theta - \phi )}}{2} = \sqrt {\frac{{4 - {a^2} - {b^2}}}{{{a^2} + {b^2}}}} $
Trick : Put $\theta = \frac{\pi }{2},\phi = {0^o}$, then $a = 1 = b$
$\tan \frac{{\theta - \phi }}{2} = 1$, which is given by $(a)$ and $(b).$
Again putting $\theta = \frac{\pi }{4} = \phi $,
we get $\tan \frac{{\theta - \phi }}{2} = 0$, which is given by $(b).$
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