Inverse Trigonometric Functions Question 192
Question: A solution of the equation $ {{\tan }^{-1}}(1+x) $ $ +{{\tan }^{-1}}(1-x) $$ =\frac{\pi }{2} $ is
[Karnataka CET 1993]
Options:
A) $ x=1 $
B) $ x=-1 $
C) $ x=0 $
D) $ x=\pi $
Show Answer
Answer:
Correct Answer: C
Solution:
$ {{\tan }^{-1}}(1+x)+{{\tan }^{-1}}(1-x)=\frac{\pi }{2} $
Therefore $ {{\tan }^{-1}}(1+x)=\frac{\pi }{2}-{{\tan }^{-1}}(1-x) $
Therefore $ {{\tan }^{-1}}(1+x)={{\cot }^{-1}}(1-x) $
Therefore $ {{\tan }^{-1}}(1+x)={{\tan }^{-1}}( \frac{1}{1-x} ) $
Therefore $ 1+x=\frac{1}{1-x}\Rightarrow 1-x^{2}=1\Rightarrow x=0 $ .
BETA
