inv-trig

Question: $\tan \left[ {{\sec }^{-1}}\sqrt{1+{{x}^{2}}} \right]=$



1) $\frac{1}{x}$
2) $x$
3) $\frac{1}{\sqrt{1+{{x}^{2}}}}$
4) $\frac{x}{\sqrt{1+{{x}^{2}}}}$
Solution: Explanation: No Explanation
Inverse trigonometric functions

Rate this question:

Average rating: (0 votes)

Previous Question Next Question