Question

If $\tan \beta =\frac{2\sin \alpha \sin \gamma }{\sin (\alpha +\gamma )}$ show that $\tan \alpha ,\tan \beta ,\tan \gamma$ are in H.P. where $\alpha \ne 0and\gamma \ne 0and\beta \ne \frac{\pi }{2}$

Answer 1

We have,

$\tan \beta =\frac{2\sin \alpha \sin \gamma }{\sin (\alpha +\gamma )}$

$\tan \beta =\frac{2\sin \alpha \sin \gamma }{\sin \alpha \cos \gamma +\cos \alpha \sin \gamma }$

$\tan \beta =\frac{\frac{2\sin \alpha \sin \gamma }{\cos \alpha \cos \gamma }}{\frac{\sin \alpha \cos \gamma +\cos \alpha \sin \gamma }{\cos \alpha \cos \gamma }}$

$\tan \beta =\frac{2\tan \alpha \tan \gamma }{\frac{\sin \alpha \cos \gamma }{\cos \alpha \cos \gamma }+\frac{\cos \alpha \sin \gamma }{\cos \alpha \cos \gamma }}$

$\tan \beta =\frac{2\tan \alpha \tan \gamma }{\frac{\sin \alpha }{\cos \alpha }+\frac{\sin \gamma }{\cos \gamma }}$

$\tan \beta =\frac{2\tan \alpha \tan \gamma }{\tan \alpha +\tan \gamma }$

So,

$\tan \alpha ,\tan \beta ,\tan \gamma$ are in $H.P$

Hence, proved.