A) \[\frac{1}{c}=\frac{1}{a}+\frac{1}{b}\]
B) \[\frac{1}{a}=\frac{1}{b}+\frac{1}{c}\]
C) \[\frac{1}{b}=\frac{1}{a}-\frac{1}{c}\]
D) \[\frac{1}{b}=\frac{1}{a}+\frac{1}{c}\]
Correct Answer: C
Solution :
\[\frac{b-c}{a}+\frac{a+c}{b}+\frac{a-b}{c}=1\] \[\Rightarrow \frac{b-c}{a}+\frac{a-b}{c}+\frac{a+c}{b}-1=0\] \[\Rightarrow \frac{b-c}{a}+\frac{a-b}{c}+\frac{a+c-b}{b}=0\] \[\Rightarrow \frac{c-b}{a}+\frac{b-a}{c}=\frac{a+c-b}{b}\] \[\Rightarrow \frac{{{c}^{2}}-bc+ab-{{a}^{2}}}{ac}=\frac{a+c-b}{b}\] \[\Rightarrow \frac{({{c}^{2}}-{{a}^{2}})-(bc-ab)}{ac}=\frac{a+c-b}{b}\] \[\Rightarrow \frac{(c-a)(c+a)-b(c-a)}{ac}\] \[=\frac{a+c-b}{b}\] \[\Rightarrow \frac{(c-a)(c+a-b)}{ac}=\frac{a+c-b}{b}\] \[\Rightarrow \frac{c-a}{ac}=\frac{1}{b}\] \[\Rightarrow \frac{a}{ac}-\frac{a}{ac}=\frac{1}{b}\] \[\Rightarrow \frac{1}{a}-\frac{1}{c}=\frac{1}{b}\]
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