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If a + b + c = 2s, what is the value of \(\frac{{{a^2} - {b^2} + {c^2} + 2ac}}{{{a^2} + {b^2} - {c^2} + 2ab}}\)
If a + b + c = 2s, what is the value of \(\frac{{{a^2} - {b^2} + {c^2} + 2ac}}{{{a^2} + {b^2} - {c^2} + 2ab}}\)
1). \(\frac{{\left( {b + s} \right)}}{{\left( {c + s} \right)}}\)
2). \(\frac{{\left( {s - b} \right)}}{{\left( {s - c} \right)}}\)
3). \(\frac{{\left( {2s - b} \right)}}{{\left( {2s - c} \right)}}\)
4). None of these
1 answers
1 vote
$(\begin{array}{l} \Rightarrow \frac{{{a^2} - {b^2} + {c^2} + 2ac}}{{{a^2} + {b^2} - {c^2} + 2ab}} = \frac{{{{\left( {a + c} \right)}^2} - {b^2}}}{{{{\left( {a + b} \right)}^2} - {c^2}}}\\ \Rightarrow \frac{{{a^2} - {b^2} + {c^2} + 2ac}}{{{a^2} + {b^2} - {c^2} + 2ab}} = \frac{{\left( {a + b + c} \right)\left( {a + c - b} \right)}}{{\left( {a + b + c} \right)\left( {a + b - c} \right)}}\\ \Rightarrow \frac{{{a^2} - {b^2} + {c^2} + 2ac}}{{{a^2} + {b^2} - {c^2} + 2ab}} = \frac{{\left( {a + c - b} \right)}}{{\left( {a + b - c} \right)}}\\ \Rightarrow \frac{{{a^2} - {b^2} + {c^2} + 2ac}}{{{a^2} + {b^2} - {c^2} + 2ab}} = \frac{{\left( {a + b + c - 2b} \right)}}{{\left( {a + b + c - 2c} \right)}}\\ \Rightarrow \frac{{{a^2} - {b^2} + {c^2} + 2ac}}{{{a^2} + {b^2} - {c^2} + 2ab}} = \frac{{2s - 2b}}{{2s - 2c}}\\ \Rightarrow \frac{{{a^2} - {b^2} + {c^2} + 2ac}}{{{a^2} + {b^2} - {c^2} + 2ab}} = \frac{{2\left( {s - b} \right)}}{{2\left( {s - c} \right)}}\\ \therefore \frac{{{a^2} - {b^2} + {c^2} + 2ac}}{{{a^2} + {b^2} - {c^2} + 2ab}} = \frac{{\left( {s - b} \right)}}{{\left( {s - c} \right)}} \end{array})$