A) 0
B) \[\pi \]
C) \[2\pi \]
D) None of these
Correct Answer: C
Solution :
[c] \[\because a-b<0,\] so \[{{\cot }^{-1}}\frac{ab+1}{a-b}={{\cot }^{-1}}b-{{\cot }^{-1}}a+\pi \] \[b-c<0,\]so, \[{{\cot }^{-1}}\frac{bc+1}{b-c}={{\cot }^{-1}}c-{{\cot }^{-1}}b+\pi \] \[c-a>0,so{{\cot }^{-1}}\frac{ca+1}{c-a}={{\cot }^{-1}}a-{{\cot }^{-1}}c\] Adding we get \[{{\cot }^{-1}}\frac{ab+1}{a-b}+{{\cot }^{-1}}\frac{bc+1}{b-c}+{{\cot }^{-1}}\frac{ca+1}{c-a}=2\pi \]
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