. IF a= x/x+z , b = y/z+x, c= z/x+y. Then Prove that 1-a/1+a + b/1+b + c/1+c = 1. Its conditional identities problem
Asked by
| 23rd Mar, 2012,
04:24: PM
Expert Answer:
x/(y+z)=a 1 + a = (x + y + z)/(y + z) a/(1+a) = x/(x+y+z)
y/(z+x)=b 1 + b = (x + y + z)/(z+x) b/(1+b) = y/(x+y+z)
z/(x+y)=c 1 + c = (x + y + z)/(x+y) c/(1+c) = z/(x+y+z)
Hence, a/1+a + b/1+b + c/1+c = 1
x/(y+z)=a 1 + a = (x + y + z)/(y + z) a/(1+a) = x/(x+y+z)
y/(z+x)=b 1 + b = (x + y + z)/(z+x) b/(1+b) = y/(x+y+z)
z/(x+y)=c 1 + c = (x + y + z)/(x+y) c/(1+c) = z/(x+y+z)
Answered by
| 23rd Mar, 2012,
04:56: PM
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