Inverse Trigonometric Functions Question 199
Question: If $ {{\tan }^{-1}}\frac{1-x}{1+x}=\frac{1}{2}{{\tan }^{-1}}x $ , then x =
Options:
A) 1
B) $ \sqrt{3} $
C) $ \frac{1}{\sqrt{3}} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
We have $ {{\tan }^{-1}}\frac{1-x}{1+x}=\frac{1}{2}{{\tan }^{-1}}x $
$ \Rightarrow {{\tan }^{-1}}[ \frac{1-\tan \theta }{1+\tan \theta } ]=\frac{1}{2}\theta $ (Putting $ x=\tan \theta ) $
$ \Rightarrow {{\tan }^{-1}}[ \frac{\tan \frac{\pi }{4}-\tan \theta }{1+\tan \frac{\pi }{4}\tan \theta } ]=\frac{\theta }{2} $
$ \Rightarrow {{\tan }^{-1}}\tan ( \frac{\pi }{4}-\theta )=\frac{\theta }{2}\Rightarrow \frac{\pi }{4}-\theta =\frac{\theta }{2} $
$ \Rightarrow \theta =\frac{\pi }{6}={{\tan }^{-1}}x\Rightarrow x=\tan \frac{\pi }{6}=\frac{1}{\sqrt{3}} $ .
BETA
