Trig2.9

From Exampleproblems

IF $\frac{m+1}{m-1}=\frac{\cos (\alpha-\beta)}{\sin (\alpha+\beta)}\,$ then prove that $m=\tan [\frac{\pi}{4}+\alpha] \tan [\frac{\pi}{4}+\beta]\,$

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

By Componendo and Dividendo, we get

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

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

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