Proving Equality in Real Numbers: The Case of a+b+c=abc

[anemone]

Let $a, b$ and $c$ be real numbers, all different from -1 and 1, such that $a+b+c=abc$. Prove that $$\frac{a^2}{1-a^2}+\frac{b^2}{1-b^2}+\frac{c^2}{1-c^2}=\frac{4abc}{(1-a^2)(1-b^2)(1-c^2)}$$

 

[kaliprasad]

Let $a, b$ and $c$ be real numbers, all different from -1 and 1, such that $a+b+c=abc$. Prove that $$\frac{a^2}{1-a^2}+\frac{b^2}{1-b^2}+\frac{c^2}{1-c^2}=\frac{4abc}{(1-a^2)(1-b^2)(1-c^2)}$$

(a + b+ c) = abc

so a = - (b+ c)/(1-bc)

if a = tan A , b = tan B , c = tan C

then tan A = - tan (B+C)

or A + B + C = npi

so 2A + 2B + 2C= 2npi

so tan 2A = tan (2npi- 2B - 2C)

or tan 2A = - tan (2B + 2C)
= (tan 2B + tan 2C)/( tan 2B tan 2C -1)

or tan 2A tan 2B tan 2C - tan 2A = tan 2B + tan 2C
or tan 2A + tan 2B + tan 2C = tan 2A tan 2B tan 2C

or
2tan A/(1-tan ^2A) + 2 tan B/(1-tan ^2 B) + 2 tanC /(1 - tan^2C) = 8 tan A tanC tanC/(( 1-tan ^2A)(1-tan ^2 B)(1- tan ^2C)

or
tan A/(1-tan ^2A) + tan B/(1-tan ^2 B) + tanC /(1 - tan^2C) = 4 tan A tanC tanC/(( 1-tan ^2A)(1-tan ^2 B)(1- tan ^2C)

or
a/(1-a^2) + b/(1-b^2) + c/(1 - c^2) = 4abc/(( 1-a^2)(1-b^2)(1- c^2))

 

[anemone]

(a + b+ c) = abc

so a = - (b+ c)/(1-bc)

if a = tan A , b = tan B , c = tan C

then tan A = - tan (B+C)

or A + B + C = npi

so 2A + 2B + 2C= 2npi

so tan 2A = tan (2npi- 2B - 2C)

or tan 2A = - tan (2B + 2C)
= (tan 2B + tan 2C)/( tan 2B tan 2C -1)

or tan 2A tan 2B tan 2C - tan 2A = tan 2B + tan 2C
or tan 2A + tan 2B + tan 2C = tan 2A tan 2B tan 2C

or
2tan A/(1-tan ^2A) + 2 tan B/(1-tan ^2 B) + 2 tanC /(1 - tan^2C) = 8 tan A tanC tanC/(( 1-tan ^2A)(1-tan ^2 B)(1- tan ^2C)

or
tan A/(1-tan ^2A) + tan B/(1-tan ^2 B) + tanC /(1 - tan^2C) = 4 tan A tanC tanC/(( 1-tan ^2A)(1-tan ^2 B)(1- tan ^2C)

or
a/(1-a^2) + b/(1-b^2) + c/(1 - c^2) = 4abc/(( 1-a^2)(1-b^2)(1- c^2))

Thanks again for participating, kaliprasad and just so you know, you and I used the same approach to solve this problem!:)