Inverse Trigonometric Functions Question 15
Question: $ {{\tan }^{-1}}\frac{1-x^{2}}{2x}+{{\cos }^{-1}}\frac{1-x^{2}}{1+x^{2}}= $
Options:
A) $ \frac{\pi }{4} $
B) $ \frac{\pi }{2} $
C) $ \pi $
D) 0
Show Answer
Answer:
Correct Answer: B
Solution:
Putting $ x=\tan \theta $
$ {{\tan }^{-1}}\frac{1-x^{2}}{2x}+{{\cos }^{-1}}\frac{1-x^{2}}{1+x^{2}} $
$ ={{\tan }^{-1}}( \frac{1-{{\tan }^{2}}\theta }{2\tan \theta } )+{{\cos }^{-1}}( \frac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta } ) $
$ ={{\tan }^{-1}}(\cot 2\theta )+{{\cos }^{-1}}(\cos 2\theta ) $
$ =\frac{\pi }{2}-2\theta +2\theta =\frac{\pi }{2} $ .
BETA
