In any ΔABC, prove that

Question:

In any ΔABC, prove that

$\left(c^{2}-a^{2}+b^{2}\right) \tan A=\left(a^{2}-b^{2}+c^{2}\right) \tan B=\left(b^{2}-c^{2}+a^{2}\right) \tan C$

 

Solution:

Need to prove: $\left(c^{2}-a^{2}+b^{2}\right) \tan A=\left(a^{2}-b^{2}+c^{2}\right) \tan B=\left(b^{2}-c^{2}+a^{2}\right) \tan C$

We know

$\tan A=\frac{a b c}{R} \frac{1}{b^{2}+c^{2}-a^{2}}-\cdots-\cdots(a)$

Similarly, $\tan B=\frac{a b c}{R} \frac{1}{c^{2}+a^{2}-b^{2}}$ and $\tan C=\frac{a b c}{R} \frac{1}{a^{2}+b^{2}-c^{2}}$

Therefore,

$\left(b^{2}+c^{2}-a^{2}\right) \tan A=\frac{a b c}{R}[$ from $(a)]$

Similarly,

$\left(c^{2}+a^{2}-b^{2}\right) \tan B=\frac{a b c}{R}$ and $\left(a^{2}+b^{2}-c^{2}\right) \tan C=\frac{a b c}{R}$

Hence we can conclude comparing above equations,

$\left(c^{2}-a^{2}+b^{2}\right) \tan A=\left(a^{2}-b^{2}+c^{2}\right) \tan B=\left(b^{2}-c^{2}+a^{2}\right) \tan C$

[Proved]

 

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