Inverse Trigonometric Functions Question 30
Question: If $ {{\tan }^{-1}}(x-1)+{{\tan }^{-1}}x+{{\tan }^{-1}}(x+1)={{\tan }^{-1}}3x $ ,then x =
Options:
A) $ \pm \frac{1}{2} $
B) $ 0,\frac{1}{2} $
C) $ 0,-\frac{1}{2} $
D) $ 0,\pm \frac{1}{2} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ {{\tan }^{-1}}(x-1)+{{\tan }^{-1}}(x)+{{\tan }^{-1}}(x+1)={{\tan }^{-1}}3x $
$ \Rightarrow {{\tan }^{-1}}(x-1)+{{\tan }^{-1}}(x)={{\tan }^{-1}}3x-{{\tan }^{-1}}(x+1) $
$ \Rightarrow {{\tan }^{-1}}[ \frac{(x-1)+x}{1-(x-1)(x)} ]={{\tan }^{-1}}[ \frac{3x-(x+1)}{1+3x(x+1)} ] $
$ \Rightarrow \frac{2x-1}{1-x^{2}+x}=\frac{2x-1}{1+3x^{2}+3x} $
$ \Rightarrow (1-x^{2}+x)(2x-1)=(1+3x^{2}+3x)(2x-1) $
On simplification $ x=0,\pm \frac{1}{2}. $
BETA
