Inverse Trigonometric Functions Question 76
Question: If $ 0<a<b<c, $ then $ {{\cot }^{-1}}( \frac{ab+1}{a-b} )+{{\cot }^{-1}}( \frac{bc+1}{b-c} )+{{\cot }^{-1}}( \frac{ca+1}{c-a} )= $
Options:
A) 0
B) $ \pi $
C) $ 2\pi $
D) None of these
Show Answer
Answer:
Correct Answer: 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 $
BETA
