If x-y-z-xyz- 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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General Solution of Trigonometric Equation

If x+y+z=xy...

Question

If $x + y + z = x y z$ , prove that $\frac{2 x}{1 - x^{2}} + \frac{2 y}{1 - y^{2}} + \frac{2 z}{1 - z^{2}} = \frac{2 x}{1 - x^{2}} \frac{2 y}{1 - y^{2}} \frac{2 z}{1 - z^{2}}$

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x + y + z = x y z

\frac{2 x}{1 - x^{2}} + \frac{2 y}{1 - y^{2}} + \frac{2 z}{1 - z^{2}} = \frac{2 x}{1 - x^{2}} \cdot \frac{2 y}{1 - y^{2}} \cdot \frac{2 z}{1 - z^{2}}

x + y + z = x y z

\frac{2 x}{1 - x^{2}} + \frac{2 y}{1 - y^{2}} + \frac{2 z}{1 - z^{2}} = \frac{r x}{1 - x^{2}} \frac{p y}{1 - y^{2}} \frac{q z}{1 - z^{2}}

p, r, q

p q r

S = \frac{2 x}{1 - x^{2}} + \frac{2 y}{1 - y^{2}} + \frac{2 z}{1 - z^{2}}

T = \frac{8 x y z}{(1 - x^{2}) (1 - y^{2}) (1 - z^{2})}

x, y, z \neq \pm 1

x + y + z = x y z

S - T

x^{3} + y^{3} + z^{3} - 3 x y z = \frac{1}{2} (x + y + z) [(x - y)^{2} + (y - z)^{2} + (z - x)^{2}]

x y + y z + z x = 1

\frac{x}{1 - x^{2}} + \frac{y}{1 - y^{2}} + \frac{z}{1 - z^{2}} = \frac{4 x y z}{(1 - x^{2}) (1 - y^{2}) (1 - z^{2})}

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