Inverse Trigonometric Functions Question 23
Question: $ \frac{1}{2}{{\cos }^{-1}}( \frac{1-x}{1+x} )= $
Options:
A) $ {{\cot }^{-1}}\sqrt{x} $
B) $ {{\tan }^{-1}}\sqrt{x} $
C) $ {{\tan }^{-1}}x $
D) $ {{\cot }^{-1}}x $
Show Answer
Answer:
Correct Answer: B
Solution:
Let $ x={{\tan }^{2}}\theta \Rightarrow \theta ={{\tan }^{-1}}\sqrt{x} $ Now, $ \frac{1}{2}{{\cos }^{-1}}( \frac{1-x}{1+x} ) $
$ =\frac{1}{2}{{\cos }^{-1}}( \frac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta } ) $
$ =\frac{1}{2}{{\cos }^{-1}}\cos 2\theta =\frac{2\theta }{2}=\theta ={{\tan }^{-1}}\sqrt{x} $ .
BETA
