A) \[a+b+c=abc\]
B) \[ab+bc+ca=abc\]
C) \[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{abc}=0\]
D) \[ab+bc+ca=a+b+c\]
E) none of the above
Correct Answer: C
Solution :
\[\because \]\[{{\tan }^{-1}}a+{{\tan }^{-1}}b+{{\tan }^{-1}}c={{\sin }^{-1}}1\] \[\Rightarrow \] \[{{\tan }^{-1}}\left( \frac{a+b+c-abc}{1-ab-bc-ca} \right)=\frac{\pi }{2}\] \[\Rightarrow \] \[\frac{a+b+c-abc}{1-ab-bc-ca}=\frac{1}{0}\] \[\Rightarrow \] \[ab+bc+ca-1=0\] \[\Rightarrow \] \[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{abc}=0\]
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