Inverse Trigonometric Functions Question 151
Question: $ {{\tan }^{-1}}( \frac{\sqrt{1+x^{2}}-1}{x} )= $
Options:
A) $ {{\tan }^{-1}}x $
B) $ \frac{1}{2}{{\tan }^{-1}}x $
C) $ 2{{\tan }^{-1}}x $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ {{\tan }^{-1}}( \frac{\sqrt{1+x^{2}}-1}{x} )={{\tan }^{-1}}[ \frac{\sqrt{1+{{\tan }^{2}}\theta }-1}{\tan \theta } ] $ (Putting $ x=\tan \theta ) $
$ ={{\tan }^{-1}}[ \frac{\sec \theta -1}{\tan \theta } ]={{\tan }^{-1}}[ \frac{1-\cos \theta }{\sin \theta } ] $
$ ={{\tan }^{-1}}[ \frac{2{{\sin }^{2}}\frac{\theta }{2}}{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}} ] $
$ ={{\tan }^{-1}}\tan \frac{\theta }{2}=\frac{\theta }{2}=\frac{1}{2}{{\tan }^{-1}}x $ .
BETA
