Question

If x + y + z = xyz, then prove that `(2x)/(1 - x^2) + (2y)/(1 - y^2) + (2z)/(1 - z^2) = (2x)/(1 - x^2) (2y)/(1 - y^2) (2z)/(1 - z^2)`

Answer 1

Taking x = tan A, y = tan B and z = tan C

`(2x)/(1 - x^2) = (2tan"A")/(1 - tan^2"A")`

= tan 2A

Similarly, `(2y)/(1 - y^2) = tan 2"B"` and `(2z)/(1 - z^2)` = tan 2C

Given x + y + x = xyz

(i.e) we are given tan A + tan B + tan C = tan A tan B tan C

⇒ A+B+C = 180°

⇒ A + B = 180° – C

Multiply 2 on both sides

⇒ 2A + 2B = 360° – 2C

2(A + B) =360° – 2C

⇒ tan(2A + 2B) = tan(360° – 2C)

= – tan 2C

(i.e) `(tan 2"A" + tan 2"B")/(1 - tan 2"A" tan 2"B")` =  – tan 2C

⇒ tan 2A + tan 2B = – tan 2C [1 – tan 2A tan 2B]

⇒ tan 2A + tan 2B = – tan 2C + tan 2A tan 2B tan 2C

⇒ tan 2A + tan 2B + tan 2C = tan 2A tan 2B tan 2C

(i.e.)  `(2x)/(1 - x^2) + (2y)/(1 - y^2) + (2z)/(1 - z^2) = (2x)/(1 - x^2) xx (2y)/(1 - y^2) xx (2z)/(1 - z^2)`