Question

If $\frac{\tan (\alpha -\beta )}{\tan \alpha }+\frac{{\sin }^2\gamma }{{\sin }^2\alpha }=1,$ then prove that tan $\gamma$ is geometric mean of tan $\alpha$ and tan $\beta$.
i.e., than $\alpha$ tan $\beta$ = ${\tan }^2\gamma$.

Answer 1

$\frac{\tan (\alpha -\beta )}{\tan \alpha }+\frac{{\sin }^2\gamma }{{\sin }^2\alpha }=1$

$\Rightarrow \frac{{\sin }^2\gamma }{{\sin }^2\alpha }=1-\frac{\tan (\alpha -\beta )}{\tan \alpha }$

$\Rightarrow \frac{{\sin }^2\gamma }{{\sin }^2\alpha }=1-\frac{\sin (\alpha -\beta )\cos \alpha }{\cos (\alpha -\beta )\sin \alpha }$

$\Rightarrow \frac{{\sin }^2\gamma }{{\sin }^2\alpha }=\frac{\sin \alpha \cos (\alpha -\beta )-\sin (\alpha -\beta )\cos \alpha }{\cos (\alpha -\beta )\sin \alpha }$

$\Rightarrow {\sin }^2\gamma =\frac{\sin (\alpha -\alpha +\beta )}{\cos (\alpha -\beta )\sin \alpha }\times {\sin }^2\alpha =\frac{\sin \beta \sin \alpha }{\cos (\alpha -\beta )}$

$\Rightarrow {\sin }^2\gamma =\frac{\sin \alpha \sin \beta }{\cos (\alpha -\beta )}$

$\Rightarrow {\csc }^2\gamma =\frac{\cos (\alpha -\beta )}{\sin \alpha \sin \beta }=\frac{\cos \alpha \cos \beta }{\sin \alpha \sin \beta }+1$

$\Rightarrow 1+{\cot }^2\gamma =\cot \alpha \cos \beta +1$

$\Rightarrow {\cot }^2\gamma =\cot \alpha \cot \beta$

$\Rightarrow {\tan }^2\gamma =\tan \alpha \tan \beta$

Hence proved