Question

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

Answer 1

$\cos \left(\alpha +\beta \right)\sin \left(\upsilon +\theta \right)=\cos \left(\alpha -\beta \right)\sin \left(\upsilon -\theta \right)$

$\frac{\cos \left(\alpha +\beta \right)}{\cos \left(\alpha -\beta \right)}=\frac{\sin \left(\upsilon -\theta \right)}{\sin \left(\upsilon +\theta \right)}$

$\frac{\cos \left(\alpha +\beta \right)+\cos \left(\alpha -\beta \right)}{\cos \left(\alpha +\beta \right)-\cos \left(\alpha -\beta \right)}=\frac{\sin \left(\upsilon -\theta \right)+\sin \left(\upsilon +\theta \right)}{\sin \left(\upsilon -\theta \right)-\sin \left(\upsilon +\theta \right)}$

$\frac{2\cos \alpha \cos \beta }{2\sin \alpha \sin \beta }=\frac{2\sin \upsilon \cos \theta }{2\cos \upsilon \sin \theta }$

$\cot \alpha \cot \beta =\frac{\cot \theta }{\cot \upsilon }$

$\frac{1}{\tan \alpha }.\frac{1}{\tan \beta }=\frac{\tan \upsilon }{\tan \theta }$

$\tan \theta =\tan \alpha \tan \beta \tan \upsilon$