Notice that if a point (a,b) is on the terminal side of θ, the following triangle is formed:
Remember then, that the length of the hypotenuse of that triangle is given by th Pythagorean Theorem:
[tex]c=\sqrt[]{a^2+b^2}[/tex]
Remember also the definitions for sinθ, secθ and tanθ:
[tex]\begin{gathered} \sin \theta=\frac{b}{c} \\ \sec \theta=\frac{c}{a} \\ \tan \theta=\frac{b}{a} \end{gathered}[/tex]
In this case, a=-5 and b=-2. Find c:
[tex]\begin{gathered} c=\sqrt[]{(-5)^2+(-2)^2} \\ =\sqrt[]{25+4} \\ =\sqrt[]{29} \end{gathered}[/tex]
Substitute the values for a, b and c to find sinθ, secθ and tanθ:
[tex]\begin{gathered} \sin \theta=-\frac{2}{\sqrt[]{29}}=-\frac{2\cdot\sqrt[]{29}}{29} \\ \sec \theta=-\frac{\sqrt[]{29}}{5} \\ \tan \theta=\frac{-2}{-5}=\frac{2}{5} \end{gathered}[/tex]
Therefore, the answers are:
[tex]\begin{gathered} \sin \theta=-\frac{2\cdot\sqrt[]{29}}{29} \\ \sec \theta=-\frac{\sqrt[]{29}}{5} \\ \tan \theta=\frac{2}{5} \end{gathered}[/tex]
