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

Consider the given equation.

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

$[\cos \alpha cos\beta -sin\alpha sin\beta ][\sin \gamma cos\theta +cos\gamma sin\theta ]=[\cos \alpha cos\beta +\sin \alpha \sin \beta ][\sin \gamma \cos \theta -\cos \gamma \sin \theta ]$

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

$2\cos \alpha cos\beta cos\gamma sin\theta =2sin\alpha sin\beta \sin \gamma cos\theta$

$\frac{sin\alpha sin\beta \sin \gamma \cos \theta }{\cos \alpha cos\beta \cos \gamma \sin \theta }=1$

$\tan \alpha \tan \beta \tan \gamma cot\theta =1$

$\tan \alpha \tan \beta \tan \gamma =\frac{1}{cot\theta }$

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

Henced, proved.