Compound angle formula Vs Componendo and dividendo

 If $ \displaystyle \frac{{m+1}}{{m-1}}=\frac{{\cos (\alpha -\beta )}}{{\sin (\alpha +\beta )}}$ , prove that $ \displaystyle \ \ \ m=\tan (\alpha +{{45}^{\circ }})\tan (\beta +{{45}^{\circ }})$


$ \displaystyle \frac{{m+1}}{{m-1}}=\frac{{\cos (\alpha -\beta )}}{{\sin (\alpha +\beta )}}$

$ \displaystyle \frac{{m+1}}{{m-1}}=\frac{{\cos \alpha \cos \beta +\sin \alpha \sin \beta }}{{\sin \alpha \cos \beta +\cos \alpha \sin \beta }}$

By Componendo and Dividendo ,

$ \displaystyle \frac{{2m}}{2}=\frac{{\cos \alpha \cos \beta +\sin \alpha \sin \beta +\sin \alpha \cos \beta +\cos \alpha \sin \beta }}{{\cos \alpha \cos \beta +\sin \alpha \sin \beta -\sin \alpha \cos \beta -\cos \alpha \sin \beta }}$

$ \displaystyle m=\frac{{\cos \beta (\sin \alpha +\cos \alpha )+\sin \beta (\sin \alpha +\cos \alpha )}}{{\cos \beta (\cos \alpha -\sin \alpha )-\sin \beta (\cos \alpha -\sin \alpha )}}$

$ \displaystyle \ \ \ =\frac{{(\sin \alpha +\cos \alpha )(\sin \beta +\cos \beta )}}{{(\cos \alpha -\sin \alpha )(\cos \beta -\sin \beta )}}$

$ \displaystyle \ \ \ =\frac{{\tan \alpha +1}}{{1-\tan \alpha }}\times \frac{{\tan \beta +1}}{{1-\tan \beta }}$

$ \displaystyle \ \ \ =\frac{{\tan \alpha +\tan {{{45}}^{\circ }}}}{{1-\tan \alpha \tan {{{45}}^{\circ }}}}\times \frac{{\tan \beta +\tan {{{45}}^{\circ }}}}{{1-\tan \beta \tan {{{45}}^{\circ }}}}$

$ \displaystyle \ \ \ =\tan (\alpha +{{45}^{\circ }})\tan (\beta +{{45}^{\circ }})$