Question
$${\frac{1}{{\left( {{{\log }_a}bc} \right) + 1}} + }$$ $${\frac{1}{{\left( {{{\log }_b}ca} \right) + 1}} + }$$ $${\frac{1}{{\left( {{{\log }_c}ab} \right) + 1}}}$$ is equal to -
Answer: Option A
$$\eqalign{
& {\text{Given}}\,\,\,{\text{Expression}} \cr
& = \frac{1}{{{{\log }_a}bc + {{\log }_a}a}} + \frac{1}{{{{\log }_b}ca + {{\log }_b}b}} + \frac{1}{{{{\log }_c}ab + {{\log }_c}c}} \cr
& = \frac{1}{{{{\log }_a}\left( {abc} \right)}} + \frac{1}{{{{\log }_b}\left( {abc} \right)}} + \frac{1}{{{{\log }_c}\left( {abc} \right)}} \cr
& = {\log _{abc}}a + {\log _{abc}}b + {\log _{abc}}c \cr
& = {\log _{abc}}\left( {abc} \right) \cr
& = 1 \cr} $$
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$$\eqalign{
& {\text{Given}}\,\,\,{\text{Expression}} \cr
& = \frac{1}{{{{\log }_a}bc + {{\log }_a}a}} + \frac{1}{{{{\log }_b}ca + {{\log }_b}b}} + \frac{1}{{{{\log }_c}ab + {{\log }_c}c}} \cr
& = \frac{1}{{{{\log }_a}\left( {abc} \right)}} + \frac{1}{{{{\log }_b}\left( {abc} \right)}} + \frac{1}{{{{\log }_c}\left( {abc} \right)}} \cr
& = {\log _{abc}}a + {\log _{abc}}b + {\log _{abc}}c \cr
& = {\log _{abc}}\left( {abc} \right) \cr
& = 1 \cr} $$
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