In a triangle $\triangle ABC$, Prove that$\frac{2}{\tan\gamma}=\frac{1}{\tan\beta}-\frac{1}{\tan\alpha}$

Question

Suppose we have triangle $\triangle ABC$ as in the figure and $AM$ be median of angle $A$. Prove that

$$\frac{2}{\tan\gamma}=\frac{1}{\tan\beta}-\frac{1}{\tan\alpha}$$

I have attempted to using law of cosines and sines as follow

$(\frac{a}{2})^2=c^2+m^2_a-2cm^2_a\cos \alpha$

$(\frac{a}{2})^2=b^2+m^2_a-2bm^2_a\cos \beta$

$\frac{a}{2\sin\alpha}=\frac{c}{\sin\gamma}$ and $\frac{a}{2\sin\beta}=\frac{b}{\sin\gamma}$ so we have $\frac{\sin\beta}{\sin\alpha}=\frac{c}{b}$

Answer 1

Drop perpendiculars $BH_1$ to $AM$ at $H_1$ and $CH_2$ to $AM$ at $H_2$. It's easy to see triangles $BH_1M$ and $CH_2M$ are congruent. Denote $BH_1=CH_2=h$, $H_1M=H_2M=d$ and $AM=m$.

Now $${1\over\tan\beta} - {1\over\tan\alpha} = {AH_2\over CH_2} - {AH_1\over BH_1} = {m+d\over h} - {m-d\over h} = {2d\over h} = 2{H_1M\over BH_1} = {2\over \tan\gamma}$$

Answer 2

Denote $\gamma'$ the angle in $\triangle AMC,$ adjacent to $\gamma.$
Clearly $$\tan\gamma'=-\tan\gamma,$$ and the identity $$\frac{2}{\tan\gamma}=\frac{1}{\tan\beta}-\frac{1}{\tan\alpha}$$ can be written $$\frac{1}{\tan\gamma}+\frac{1}{\tan\alpha}=\frac{1}{\tan\gamma'}+\frac{1}{\tan\beta}$$ or equivalently $$\frac{\sin(\alpha + \gamma)}{\sin\alpha \sin\gamma}=\frac{\sin(\beta+ \gamma')}{\sin\beta\sin\gamma'}.$$ This identity can be proved with the use of the law of sines in triangles $ABM$ and $AMC,$ respectively. We benefit from the fact that the triangles have two sides of equal lengths, and that $\sin\gamma'=\sin\gamma.$ $$\frac{\sin(\alpha + \gamma)}{|AM|}=\frac{\sin\alpha}{|BM|},\; \frac{\sin(\beta+ \gamma')}{|AM|}=\frac{\sin\beta}{|BM|}$$