If xy-yz-zx-1- then x1-x2-y1-y2-z1-z2 is equal to

The correct option is D 2/√((1+x^2)(1+y^2)(1+z^2)) Let x=tanA2 y=tanB2 and z=tanC2 .Where A+B+C=π .Thence tanA2.tanB2+tanB2.tanC ...

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Standard XII

Mathematics

Integration of Irrational Algebraic Fractions - 2

If xy+yz+zx...

Question

If $x y + y z + z x = 1$ , then $\frac{x}{1 + x^{2}} + \frac{y}{1 + y^{2}} + \frac{z}{1 + z^{2}}$ is equal to

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

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

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

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

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Solution

The correct option is D $\frac{2}{\sqrt{(1 + x^{2}) (1 + y^{2}) (1 + z^{2})}}$
Let $x = tan \frac{A}{2}$ $y = tan \frac{B}{2}$ and $z = tan \frac{C}{2}$ .
Where $A + B + C = π$ .
Thence
$tan \frac{A}{2} . tan \frac{B}{2} + tan \frac{B}{2} . tan \frac{C}{2} + tan \frac{C}{2} tan \frac{A}{2} = 1$ .... (property).
Now
$\frac{x}{1 + x^{2}} + \frac{y}{1 + y^{2}} + \frac{z}{1 + z^{2}}$
$= \frac{tan (\frac{A}{2})}{1 + {tan}^{2} (\frac{A}{2})} + \frac{tan (\frac{B}{2})}{1 + {tan}^{2} (\frac{B}{2})} + \frac{tan (\frac{C}{2})}{1 + {tan}^{2} (\frac{C}{2})}$
$= \frac{1}{2} [sin A + sin B + sin C]$
$= \frac{1}{2} [4 cos \frac{C}{2} . cos \frac{B}{2} . cos \frac{A}{2}]$
$= 2 cos \frac{C}{2} . cos \frac{B}{2} . cos \frac{A}{2}$
$= \frac{2}{\sqrt{(1 + tan \frac{A}{2})^{2} (1 + tan \frac{B}{2})^{2} (1 + tan \frac{C}{2})^{2}}}$
$= \frac{2}{\sqrt{(1 + x^{2}) (1 + y^{2}) (1 + z^{2})}}$

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Similar questions

Q. If

x y + y z + z x = 1

, the prove that

\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}}

Q. If

x y + y z + z x = 1

, prove that

\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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\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})} .

If $∣ ∣ ∣ ∣ ∣ \begin{matrix} x^{n} & x^{x + 2} & x^{x + 4} y^{n} & y^{n + 2} & y^{n + 4} z^{n} & z^{n + 2} & z^{n + 4} \end{matrix} ∣ ∣ ∣ ∣ ∣ = (\frac{1}{y^{2}} - \frac{1}{x^{2}}) (\frac{1}{z^{2}} - \frac{1}{y^{2}}) (\frac{1}{x^{2}} - \frac{1}{z^{2}})$ then n=

Q. Statement

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x y + y z + z x = 1

where

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

, then

\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})}}

Statement

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If xy+yz+zx=1, then x1+x2+y1+y2+z1+z2 is equal to

If $x y + y z + z x = 1$ , then $\frac{x}{1 + x^{2}} + \frac{y}{1 + y^{2}} + \frac{z}{1 + z^{2}}$ is equal to