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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)`
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Solution
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)`
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