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If $A,B,C$ are angles of a triangle then prove that $$\frac{\operatorname{tan}A}{\operatorname{tan}B}=\frac{c^2+a^2-b^2}{c^2+b^2-a^2}$$ where $\angle A$ is opposite to side $a$ , $\angle B$ to side $b$ and $\angle C$ to side $c$.

I wasn't having a clue of proceeding. So, I applied another method. If the above expression is true then this expression must also be true $$\frac{\operatorname{tan}A+\operatorname{tan}B}{\operatorname{tan}A-\operatorname{tan}B}=\frac{c^2+a^2-b^2+c^2+b^2-a^2}{c^2+a^2-b^2-c^2-b^2+a^2}$$ $$\frac{\operatorname{tan}A+\operatorname{tan}B}{\operatorname{tan}A-\operatorname{tan}B}=\frac{c^2}{a^2-b^2}$$ Now, I am stuck here.

I applied sine rule and got the result $$\frac{c^2+a^2-b^2}{c^2+b^2-a^2}=\frac{\operatorname{sin}^2C+\operatorname{sin}^2A-\operatorname{sin}^2B}{\operatorname{sin}^2C+\operatorname{sin}^2B-\operatorname{sin}^2A}$$

Any help is greatly appreciated.

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    $\begingroup$ Can you think of a trigonometric rule which involves the squares of the sides? $\endgroup$ Sep 5, 2022 at 14:56

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Use the sine and cosine rules: $$\tan A =\frac{\sin A}{\cos A}, \tan B=\frac{\sin B}{\cos B}$$ $$\text{Sine Rule:}\quad \sin A=\frac{a}{2R}, \sin B=\frac{b}{2R}$$ $$\text{Cosine Rule:}\quad \cos A=\frac{c^2+b^2-a^2}{2bc}, \cos B=\frac{c^2+a^2-b^2}{2ac}$$ Thus, $$\tan A=\frac{abc}{R}\frac{1}{c^2+b^2-a^2}, \tan B= \frac{abc}{R}\frac{1}{c^2+a^2-b^2} $$ Divide. Hence proved.

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    $\begingroup$ The OP may know the weaker version of the Sine rule: $\frac{\sin A}{a}=\frac{sin B}{b}=\frac{\sin C}{c}$, thus ideally this proof would just call that coefficient of proportionality $k$ (or something) - it will cancel out anyways, and we don't need to know that $k=\frac{1}{2R}$, $\endgroup$
    – user700480
    Sep 5, 2022 at 15:50

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