(b) चूँकि $\tan \theta = \frac{{x\,\sin \,\phi }}{{1 – x\,\cos \,\,\phi }}$

$ \Rightarrow \,\,\frac{1}{x}\tan \theta – \tan \theta \,\,\cos \phi = \sin \,\phi $ 

$ \Rightarrow \,\,\frac{1}{x} = \frac{{\sin \,\phi + \cos \,\,\phi \,\tan \,\theta }}{{\tan \,\theta }}$

एवं $\tan \,\phi = \frac{{y\,\sin \,\theta }}{{1 – y\,\cos \,\theta }}$

$ \Rightarrow \tan \,\phi \, = \frac{{\sin \,\theta }}{{\frac{1}{y} – \cos \,\theta }}$

$ \Rightarrow \,\,\frac{1}{y}\tan \,\phi – \tan \,\phi \cos \theta = \sin \theta $

$ \Rightarrow \,\,\frac{1}{y}\tan \,\phi \, = \sin \,\theta + \tan \,\phi \,\cos \theta $

$\therefore \,\,\,\,\frac{1}{y} = \frac{{\sin \,\theta + \tan \,\phi \,\cos \theta }}{{\tan \,\phi }}$

अब $\frac{x}{y} = \left[ {\frac{{\tan \,\theta }}{{\sin \,\phi + \cos \,\phi \,\tan \,\theta }}} \right] \times \left[ {\frac{{\sin \,\theta + \tan \,\phi \,\cos \,\theta }}{{\tan \,\phi }}} \right]$

$ = \frac{{\tan \,\theta }}{{\tan \,\phi }}\,\left[ {\frac{{\sin \,\theta + \cos \,\theta \frac{{\sin \varphi }}{{\cos \phi }}}}{{\sin \phi + \cos \phi \frac{{\sin \theta }}{{\cos \theta }}}}} \right] $

$= \frac{{\tan \theta \,\,\cos \theta }}{{\tan \,\phi \,\cos \phi }} = \frac{{\sin \theta }}{{\sin \phi }}$

वैकल्पिक​ : $x\,\sin \,\phi = \tan \,\theta – x\,\cos \,\phi \,\tan \,\theta $

$ \Rightarrow \,x = \frac{{\tan \,\theta }}{{\sin \,\phi + \cos \,\phi \,\tan \,\theta }}$

$ = \frac{{\sin \,\theta }}{{\cos \,\theta \sin \,\phi + \cos \,\phi \,\sin \,\theta }} $

$= \frac{{\sin \,\theta }}{{\sin \,(\theta + \phi )}}$

इसी प्रकार, $y = \frac{{\sin \,\phi }}{{\sin \,(\theta + \phi )}}$;

$\therefore \,\,\frac{x}{y} = \frac{{\sin \theta }}{{\sin \phi }}.$