JEE Main On 8 April 2017 Question 4
Question: The value of $ {{\tan }^{-1}}[ \frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}} ],|x|<\frac{1}{2},x\ne 0, $ is equal to: [JEE Online 08-04-2017]
Options:
A) $ \frac{\pi }{4}+\frac{1}{2}{{\cos }^{-1}}x^{2} $
B) $ \frac{\pi }{4}-{{\cos }^{-1}}x^{2} $
C) $ \frac{\pi }{4}-\frac{1}{2}{{\cos }^{-1}}x^{2} $
D) $ \frac{\pi }{4}+{{\cos }^{-1}}x^{2} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ x^{2}=\cos 2\theta ;\theta =\frac{1}{2}\cos x^{2} $
$ {{\tan }^{-1}}[ \frac{\sqrt{1+\cos 2\theta }+\sqrt{1-\cos 2\theta }}{\sqrt{1+\cos 2\theta }-\sqrt{1-\cos 2\theta }} ] $
$ {{\tan }^{-1}}[ \frac{1+\tan \theta }{1-\tan \theta } ] $
$ ={{\tan }^{-1}}[ \tan ( \frac{\pi }{4}+\theta ) ] $
$ =\frac{\pi }{4}+\frac{1}{2}\cos ^{-1} x^{2} $
