| If \[x+y+z=0,\]then the value of \[\frac{1}{{{x}^{2}}+{{y}^{2}}-{{z}^{2}}}+\frac{1}{{{x}^{2}}+{{z}^{2}}-{{y}^{2}}}+\frac{1}{{{y}^{2}}+{{z}^{2}}-{{x}^{2}}}\]is [SSC (CGL) 2014] |
A) \[-\,\,2\]
B) \[-\frac{1}{2}\]
C) \[0\]
D) \[\frac{1}{2}\]
Correct Answer: C
Solution :
| Expression is \[\frac{1}{{{x}^{2}}+{{y}^{2}}-{{z}^{2}}}\]\[+\frac{1}{{{x}^{2}}+{{z}^{2}}-{{y}^{2}}}+\frac{1}{{{y}^{2}}+{{z}^{2}}-{{x}^{2}}}\] |
| Given that, \[x+y+z=0\] |
| \[\therefore \]\[x=-\,\,(y+z),\]\[y=-\,\,(x+z)\]and \[z=-\,\,(x+y)\] |
| Putting the values of x, y and z in the expression |
| \[=\frac{1}{{{x}^{2}}+{{y}^{2}}-{{\{-\,\,(x+y)\}}^{2}}}+\frac{1}{{{x}^{2}}+{{z}^{2}}-{{\{-\,\,(x+z)\}}^{2}}}\]\[+\frac{1}{{{y}^{2}}+{{z}^{2}}-{{\{-\,\,(y+z)\}}^{2}}}\] |
| \[=\frac{1}{{{x}^{2}}+{{y}^{2}}-{{x}^{2}}-{{y}^{2}}-2xy}+\] |
| \[\frac{1}{{{x}^{2}}+{{z}^{2}}-{{x}^{2}}-{{z}^{2}}-2xz}+\frac{1}{{{y}^{2}}+{{z}^{2}}-{{y}^{2}}-{{z}^{2}}-2yz}\] |
| \[=\frac{-1}{2xy}-\frac{1}{2xz}-\frac{1}{2yz}=-\frac{1}{2}\left( \frac{1}{xy}+\frac{1}{xz}+\frac{1}{yz} \right)\] |
| \[=-\frac{1}{2}\left( \frac{x+y+z}{xyz} \right)=-\frac{1}{2}\left( \frac{0}{xyz} \right)\] |
| \[[\because x+y+z=0\,\,\text{(given) }\!\!]\!\!\text{ }\] |
| \[=0\] |
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