Question

If $\cos (\alpha +\beta )sin(\gamma +\theta )=cos(\alpha -\beta )sin(\gamma -\theta )$ . show that $\tan \theta =\tan \alpha \tan \beta \tan \gamma$.

Answer 1

$\cos (\alpha +\beta )\sin (\gamma +\theta )=\cos (\alpha -\beta )\sin (\gamma -\theta )$

$⟹\frac{\cos (\alpha +\beta )}{\cos (\alpha -\beta )}=\frac{\sin (\gamma -\theta )}{\sin (\gamma +\theta )}$

$⟹\frac{\cos \alpha \cos \beta -\sin \alpha \sin \beta }{\cos \alpha \cos \beta +\sin \alpha \sin \beta }=\frac{\sin \gamma \cos \theta -\cos \gamma \sin \theta }{\sin \gamma \cos \theta +\cos \gamma \sin \theta }$

$⟹\frac{1-\tan \alpha \tan \beta }{1+\tan \alpha \tan \beta }=\frac{\tan \gamma -\tan \theta }{\tan \gamma +\tan \theta }$

Applying compoundo and dividendo rule

$⟹\frac{1-\tan \alpha \tan \beta +1+\tan \alpha \tan \beta }{1-\tan \alpha \tan \beta -1-\tan \alpha \tan \beta }=\frac{\tan \gamma -\tan \theta +\tan \gamma +\tan \theta }{\tan \gamma -\tan \theta -\tan \gamma -\tan \theta }$

$⟹\frac{2}{-2\tan \alpha \tan \beta }=\frac{2\tan \gamma }{-2\tan \theta }$

$⟹\tan \theta =\tan \alpha \tan \beta \tan \gamma$