$${{\tan }^{-1}}\,\left[ \frac{\sqrt{1+{{x}^{2}}}+\sqrt{1-{{x}^{2}}}}{\sqrt{1+{{x}^{2}}}-\sqrt{1-{{x}^{2}}}} \right]=$$

Question

TuteeHUB · Accepted Answer

$$\frac{\pi }{4}+{{\cos }^{-1}}{{x}^{2}}$$

TuteeHUB · Answer

$$\frac{\pi }{4}+\frac{1}{2}{{\cos }^{-1}}{{x}^{2}}$$

TuteeHUB · Answer

$$\frac{\pi }{4}+\frac{1}{2}{{\cos }^{-1}}x$$

TuteeHUB · Answer

$$\frac{\pi }{4}-\frac{1}{2}{{\cos }^{-1}}{{x}^{2}}$$

1.	\[{{\tan }^{-1}}\,\left[ \frac{\sqrt{1+{{x}^{2}}}+\sqrt{1-{{x}^{2}}}}{\sqrt{1+{{x}^{2}}}-\sqrt{1-{{x}^{2}}}} \right]=\]
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.	\[\frac{\pi }{4}-\frac{1}{2}{{\cos }^{-1}}{{x}^{2}}\]
Answer» B. \[\frac{\pi }{4}+{{\cos }^{-1}}{{x}^{2}}\]

\[{{\tan }^{-1}}\,\left[ \frac{\sqrt{1+{{x}^{2}}}+\sqrt{1-{{x}^{2}}}}{\sqrt{1+{{x}^{2}}}-\sqrt{1-{{x}^{2}}}} \right]=\]