In any ΔABC, prove that (c2 – a2 + b2) tan A = (a2 – b2 + c2) ta

1.	In any ΔABC, prove that (c2 – a2 + b2) tan A = (a2 – b2 + c2) tan B = (b2 – c2 + a2) tan C
Answer» To prove (c² – a² + b²) tan A = (a² – b² + c²) tan B = (b² – c² + a²⁾ tan C We know tan A = \(\frac{abc}R\)\(\frac{1}{b^2+c^2-a^2}\).......(a) Similarity, tan B = \(\frac{abc}R\)\(\frac{1}{b^2+c^2-a^2}\) and tanC = \(\frac{abc}R\)\(\frac{1}{b^2+c^2-a^2}\) Therefore, (c² - a² + b²) tanA = (a² - b² + c²) tanB = (b² - c² +a²) tanc = \(\frac{abc}R\) Hence we can conclude comparing above equations. (c²– a² + b²) tanA = (a² – b² + c²) tanB = (b² – c² + a²) tanC [Proved]

In any ΔABC, prove that (c2 – a2 + b2) tan A = (a2 – b2 + c2) tan B = (b2 – c2 + a2) tan C

Answer»

To prove

(c² – a² + b²) tan A = (a² – b² + c²) tan B = (b² – c² + a²⁾ tan C

We know

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

Similarity, tan B = \(\frac{abc}R\)\(\frac{1}{b^2+c^2-a^2}\) and tanC = \(\frac{abc}R\)\(\frac{1}{b^2+c^2-a^2}\)

Therefore,

(c² - a² + b²) tanA = (a² - b² + c²) tanB = (b² - c² +a²) tanc = \(\frac{abc}R\)

Hence we can conclude comparing above equations.

(c²– a² + b²) tanA = (a² – b² + c²) tanB = (b² – c² + a²) tanC

[Proved]