The value of x-y-y-x y-z-z-y z-x-x-z - 1-x2-1-y2 1-y2-1-z2 1-z2-1-x2 is

The correct option is A x^2y^2z^2 (x/y y/x )÷ (1/x^2 1/y^2 )= xy etc. ∴ Given expression =( xy)( yz)( xz) = x^2y^2z^2 ...

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Euler's Representation

The value of ...

Question

The value of $\frac{(\frac{x}{y} - \frac{y}{x}) (\frac{y}{z} - \frac{z}{y}) (\frac{z}{x} - \frac{x}{z})}{(\frac{1}{x^{2}} - \frac{1}{y^{2}}) (\frac{1}{y^{2}} - \frac{1}{z^{2}}) (\frac{1}{z^{2}} - \frac{1}{x^{2}})}$ is

- x^{2} y^{2} z^{2}

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x^{2} y^{2} z^{2}

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xyz

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Solution

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Similar questions

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})}

x y + y z + z x = 1

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

I f 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})} .

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

1 :

x y + y z + z x = 1

x, y, z \in R^{+}

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

2 :

A B C

sin 2 A + sin 2 B + sin 2 C = 4 sin A sin B sin C

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