Inverse Trigonometric Functions Question 181
Question: $ {{\tan }^{-1}}[ \frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}} ]= $
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 $
D) $ \frac{\pi }{4}-\frac{1}{2}{{\cos }^{-1}}x^{2} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ {{\tan }^{-1}}[ \frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}-\sqrt{1-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 }} ] $ (Putting $ x^{2}=\cos 2\theta \Rightarrow \theta =\frac{1}{2}{{\cos }^{-1}}x^{2}) $ = $ {{\tan }^{-1}}[ \frac{\sqrt{2}\cos \theta +\sqrt{2}\sin \theta }{\sqrt{2}\cos \theta -\sqrt{2}\sin \theta } ] $
$ ={{\tan }^{-1}}[ \frac{1+\tan \theta }{1-\tan \theta } ]={{\tan }^{-1}}[ \frac{\tan \frac{\pi }{4}+\tan \theta }{1-\tan \frac{\pi }{4}\tan \theta } ] $
$ ={{\tan }^{-1}}\tan ( \frac{\pi }{4}+\theta )=\frac{\pi }{4}+\theta =\frac{\pi }{4}+\frac{1}{2}{{\cos }^{-1}}x^{2} $ .
BETA
