So, here we have the point:
[tex](-\frac{11}{61},\frac{60}{61})[/tex]
This is situated at the quadrant II.
Remember that sin(a) is a relation between the opposite side of the angle a and the hypotenuse of the triangle.
To find the hypotenuse, we apply the Pythagorean Theorem:
[tex]\begin{gathered} h=\sqrt[]{(\frac{60}{61})^2+(\frac{-11}{61})^2} \\ h=1 \end{gathered}[/tex]
Notice that as this point is in the unit circle, the hypotenuse is 1.
Now,
[tex]\sin (\theta)=\frac{60}{61}[/tex]
And,
[tex]sec(\theta)=\frac{1}{\cos (\theta)}=\frac{-61}{11}[/tex][tex]\text{tan(}\theta)=\frac{\frac{60}{61}}{\frac{-11}{61}}=\frac{-60}{11}[/tex]
