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) \[\frac{\pi }{4}-\frac{1}{2}{{\cos }^{-1}}{{x}^{2}}\]
Correct Answer: A
Solution :
\[{{\tan }^{-1}}\left[ \frac{\sqrt{1+{{x}^{2}}}+\sqrt{1-{{x}^{2}}}}{\sqrt{1+{{x}^{2}}}-\sqrt{1-{{x}^{2}}}} \right]\] \[={{\tan }^{-1}}\left[ \frac{\sqrt{1+\cos 2\theta }+\sqrt{1-\cos 2\theta }}{\sqrt{1+\cos 2\theta }-\sqrt{1-\cos 2\theta }} \right]\](Putting \[{{x}^{2}}=\cos 2\theta \Rightarrow \theta =\frac{1}{2}{{\cos }^{-1}}{{x}^{2}})\] = \[{{\tan }^{-1}}\left[ \frac{\sqrt{2}\cos \theta +\sqrt{2}\sin \theta }{\sqrt{2}\cos \theta -\sqrt{2}\sin \theta } \right]\] \[={{\tan }^{-1}}\left[ \frac{1+\tan \theta }{1-\tan \theta } \right]={{\tan }^{-1}}\left[ \frac{\tan \frac{\pi }{4}+\tan \theta }{1-\tan \frac{\pi }{4}\tan \theta } \right]\] \[={{\tan }^{-1}}\tan \left( \frac{\pi }{4}+\theta \right)=\frac{\pi }{4}+\theta =\frac{\pi }{4}+\frac{1}{2}{{\cos }^{-1}}{{x}^{2}}\].
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