Let $a,b,c$ be the side lengths...find $\frac{\tan C}{\tan A}+\frac{\tan C}{\tan B}$.

Question

Let $a,b,c$ be the side lengths of an acute triangle $\triangle ABC$, respectively. If $\frac{b}{a}+\frac{a}{b}=6 \cos C$, determine $$\frac{\tan C}{\tan A}+\frac{\tan C}{\tan B}.$$

Answer 1

Using the theorem of cosines we get $$\frac{b}{a}+\frac{a}{b}=\frac{6(a^2+b^2-c^2)}{2ab}$$ from here we obtain $$c^2=\frac{2}{3}(a^2+b^2)$$ so we get $$ \cos(\gamma)=\frac{a^2+b^2}{2ab}$$ Now we use that $$\frac{a}{b}+\frac{b}{a}\geq 2$$ so $$\cos(\gamma)=\frac{1}{6}\left(\frac{a}{b}+\frac{b}{a}\right)\geq \frac{1}{3}$$ since $$\cos(\gamma)\le 1$$ we get …? With $$\cos(\gamma)$$ we get $$\sin(\gamma)$$

Answer 2

First,${a\over b}+{b\over a}=6\cos C $ is equal to $a^2+b^2=6ab\cos C$.

So $c^2=a^2+b^2-2ab\cos C=4ab\cos C$.

So $\sin^2C=4\sin A\sin B\cos C$.

Thus $${\tan C \over \tan A}+{\tan C \over \tan B}=\tan C({{\tan A+\tan B}\over \tan A\tan B})={\sin C \over \cos C}*{{\sin A\cos B+\sin B\cos A}\over \sin A\sin B}={{\sin^2C}\over \sin A\sin B\cos C}=4$$