Question

If $ab>-1,bc>-1$ and $ca>-1$, then the value of ${\cot }^{-1}\left(\frac{ab+1}{a-b}\right)+{\cot }^{-1}\left(\frac{bc+1}{b-c}\right)+{\cot }^{-1}\left(\frac{ca+1}{c-a}\right)$ is

• A. $-1$
• B. ${\cot }^{-1}(a+b+c)$
• C. ${\cot }^{-1}\left(abc\right)$
• D. $0$
• E. ${\tan }^{-1}(a+b+c)$

Answer 1

The correct option is D $0$

Given, $ab>-1,bc>-1$ and $ca>-1$, then
${\cot }^{-1}\left(\frac{ab+1}{a-b}\right)+{\cot }^{-1}\left(\frac{bc+1}{b-c}\right)+{\cot }^{-1}\left(\frac{ca+1}{c-a}\right)$
$=({\cot }^{-1}b-{\cot }^{-1}a)+({\cot }^{-1}c-{\cot }^{-1}b)+({\cot }^{-1}a-{\cot }^{-1}c)$
$=0$.