Answer
$${\sin ^{ - 1}}\left( {\frac{x}{6}} \right),{\text{ }}{\tan ^{ - 1}}\left( {\frac{x}{{\sqrt {36 - {x^2}} }}} \right),{\text{ }}{\sec ^{ - 1}}\left( {\frac{6}{{\sqrt {36 - {x^2}} }}} \right)$$
Work Step by Step
$$\eqalign{
& {\text{From the given triangle we can calculate the adjacent side using}} \cr
& {\text{the pyhtagorean theorem:}} \cr
& {\text{adjacent side}} = \sqrt {{{\left( 6 \right)}^2} - {x^2}} \cr
& {\text{adjacent side}} = \sqrt {36 - {x^2}} \cr
& \cr
& {\text{Calculating }}\sin \theta ,{\text{ tan }}\theta {\text{ and sec}}\theta \cr
& {\text{sin}}\theta = \frac{{{\text{opposite side}}}}{{{\text{hypotenuse}}}} \cr
& {\text{sin}}\theta = \frac{x}{6} \Rightarrow \theta = {\sin ^{ - 1}}\left( {\frac{x}{6}} \right) \cr
& {\text{tan}}\theta = \frac{{{\text{opposite side}}}}{{{\text{adjacent side}}}} \cr
& \tan \theta = \frac{x}{{\sqrt {36 - {x^2}} }} \Rightarrow \theta = {\tan ^{ - 1}}\left( {\frac{x}{{\sqrt {36 - {x^2}} }}} \right) \cr
& {\text{sec}}\theta = \frac{{{\text{hypotenuse}}}}{{{\text{adjacent side}}}} \cr
& \sec \theta = \frac{6}{{\sqrt {36 - {x^2}} }} \Rightarrow \theta = {\sec ^{ - 1}}\left( {\frac{6}{{\sqrt {36 - {x^2}} }}} \right) \cr} $$
