\(\frac{1}{\sqrt{r}}\)
\(\sqrt{r}\)
\(\sqrt{r^2 + 1}\)
\(\sqrt{r^2 - 1}\)
Correct answer is D
cos xo = \(\frac{1}{r}\); \(\sqrt{r^2 - 1}\)
By Pythagoras r2 = 12 + x2 - 1
x = \(\sqrt{r^2 - 1}\)
tan xo = \(\sqrt{r^2 - 1}\)
= \(\sqrt{r^2 - 1}\)