Inverse Trigonometric Functions Question 106
Question: If $ {{\sin }^{-1}}( \frac{2a}{1+a^{2}} )-{{\cos }^{-1}}( \frac{1-b^{2}}{1+b^{2}} )={{\tan }^{-1}}( \frac{2x}{1-x^{2}} ), $ then what is the value of x?
Options:
A) $ a/b $
B) $ ab $
C) $ b/a $
D) $ \frac{a-b}{1+ab} $
Show Answer
Answer:
Correct Answer: D
Given, $ {{\sin }^{-1}}( \frac{2a}{1+a^{2}} )-{{\cos }^{-1}}( \frac{1-b^{2}}{1+b^{2}} )=ta{b^{-1}}( \frac{2x}{1-x^{2}} ) $
$ \therefore 2{{\tan }^{-1}}a-2{{\tan }^{-1}}b=2{{\tan }^{-1}}x $
$ \Rightarrow {{\tan }^{-1}}a-{{\tan }^{-1}}b={{\tan }^{-1}}x $
$ \Rightarrow {{\tan }^{-1}}( \frac{a-b}{1+ab} )={{\tan }^{-1}}x $
$ \Rightarrow x=\frac{a-b}{1+ab} $
BETA
