If xy+yz+zx=1 then x1+x2+y1+y2+z1+z2=

Let x=tan⁡A2y=tan⁡B2 and z=tan⁡C2. Where A+B+C=π. ∴tan⁡A2⋅tan⁡B2+tan⁡B2⋅tan⁡C2+tan⁡C2tan⁡A2=1 x1+x2+y1+y2+z1+z2=tan⁡A21+tan2⁡A2+tan⁡B21+tan2⁡B2+tan⁡C21+tan2⁡C2=12[sin⁡A+sin⁡B+sin⁡C]=124cos⁡C2⋅cos⁡B2⋅cos⁡A2=2cos⁡C2⋅cos⁡B2⋅cos⁡A2 =21+tan⁡A221+tan⁡B221+tan⁡C22=21+x21+y21+z2

If xy+yz+zx=1 then x1+x2+y1+y2+z1+z2=

Let x=tan⁡A2y=tan⁡B2 and z=tan⁡C2. Where A+B+C=π. ∴tan⁡A2⋅tan⁡B2+tan⁡B2⋅tan⁡C2+tan⁡C2tan⁡A2=1 x1+x2+y1+y2+z1+z2=tan⁡A21+tan2⁡A2+tan⁡B21+tan2⁡B2+tan⁡C21+tan2⁡C2=12[sin⁡A+sin⁡B+sin⁡C]=124cos⁡C2⋅cos⁡B2⋅cos⁡A2=2cos⁡C2⋅cos⁡B2⋅cos⁡A2 =21+tan⁡A221+tan⁡B221+tan⁡C22=21+x21+y21+z2

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If $x y + y z + z x = 1$ then $\frac{x}{1 + x^{2}} + \frac{y}{1 + y^{2}} + \frac{z}{1 + z^{2}} =$

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

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

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

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

answer is C.

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Detailed Solution

$Let x = \tan \frac{A}{2} y = \tan \frac{B}{2} and z = \tan \frac{C}{2} . Where A + B + C = π . ∴ \tan \frac{A}{2} \cdot \tan \frac{B}{2} + \tan \frac{B}{2} \cdot \tan \frac{C}{2} + \tan \frac{C}{2} \tan \frac{A}{2} = 1 \begin{array}{l} \frac{x}{1 + x^{2}} + \frac{y}{1 + y^{2}} + \frac{z}{1 + z^{2}} \\ = \frac{\tan (\frac{A}{2})}{1 + \tan^{2} (\frac{A}{2})} + \frac{\tan (\frac{B}{2})}{1 + \tan^{2} (\frac{B}{2})} + \frac{\tan (\frac{C}{2})}{1 + \tan^{2} (\frac{C}{2})} \\ = \frac{1}{2} [\sin A + \sin B + \sin C] \\ = \frac{1}{2} [4 \cos \frac{C}{2} \cdot \cos \frac{B}{2} \cdot \cos \frac{A}{2}] \\ = 2 \cos \frac{C}{2} \cdot \cos \frac{B}{2} \cdot \cos \frac{A}{2} \begin{array}{l} = \frac{2}{\sqrt{{(1 + \tan \frac{A}{2})}^{2} {(1 + \tan \frac{B}{2})}^{2} {(1 + \tan \frac{C}{2})}^{2}}} \\ = \frac{2}{\sqrt{(1 + x^{2}) (1 + y^{2}) (1 + z^{2})}} \end{array} \end{array}$