Inverse Trigonometric Functions Question 237
Question: $ \sin { {{\tan }^{-1}}( \frac{1-x^{2}}{2x} )+{{\cos }^{-1}}( \frac{1-x^{2}}{1+x^{2}} ) } $ is equal to
[Kurukshetra CEE 2001]
Options:
A) 0
B) 1
C) $ \sqrt{2} $
D) $ \frac{1}{\sqrt{2}} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ \sin [ {{\tan }^{-1}}( \frac{1-x^{2}}{2x} )+{{\cos }^{-1}}( \frac{1-x^{2}}{1+x^{2}} ) ] $
Putting $ x=\tan \theta $
we get, $ \sin [ {{\tan }^{-1}}( \frac{1-{{\tan }^{2}}\theta }{2\tan \theta } )+{{\cos }^{-1}}( \frac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta } ) ] $ = $ \sin [{{\tan }^{-1}}(\cot 2\theta )+{{\cos }^{-1}}(\cos 2\theta )] $
= $ \sin [{{\tan }^{-1}}\tan (\pi /2-2\theta )+{{\cos }^{-1}}\cos 2\theta ] $ = $ \sin \frac{\pi }{2}=1 $ .
BETA
