Inverse Trigonometric Functions Question 201
Question: $ {{\tan }^{-1}}\frac{a-b}{1+ab}+{{\tan }^{-1}}\frac{b-c}{1+bc}= $
Options:
A) $ {{\tan }^{-1}}a-{{\tan }^{-1}}b $
B) $ {{\tan }^{-1}}a-{{\tan }^{-1}}c $
C) $ {{\tan }^{-1}}b-{{\tan }^{-1}}c $
D) $ {{\tan }^{-1}}c-{{\tan }^{-1}}a $
Show Answer
Answer:
Correct Answer: B
Solution:
$ {{\tan }^{-1}}( \frac{a-b}{1+ab} )+{{\tan }^{-1}}( \frac{b-c}{1+bc} ) $
$ ={{\tan }^{-1}}(a)-{{\tan }^{-1}}(b)+{{\tan }^{-1}}(b)-{{\tan }^{-1}}(c) $
$ ={{\tan }^{-1}}(a)-{{\tan }^{-1}}(c) $ .
BETA
