Answer:
[tex]sin(t) =\frac{\sqrt{7}}{4}[/tex]
Step-by-step explanation:
By definition we know that
[tex]sec(t) = \frac{1}{cos(t)}[/tex]
and
[tex]cos ^ 2(t) = 1-sin ^ 2(t)[/tex]
As [tex]sec(t) = -\frac{4}{3}[/tex]
Then
[tex]sec(t) = -\frac{4}{3}\\\\\frac{1}{cos(t)} =-\frac{4}{3}\\\\cos(t) = -\frac{3}{4}[/tex]
Now square both sides of the equation:
[tex]cos^2(t) = (-\frac{3}{4})^2[/tex]
[tex]cos^2(t) = \frac{9}{16}\\\\[/tex]
[tex]1-sin^2(t) =\frac{9}{16}\\\\sin^2(t) =1-\frac{9}{16}\\\\sin^2(t) =\frac{7}{16}\\\\sin(t) =\±\sqrt{\frac{7}{16}}[/tex]
In the second quadrant sin (t) is positive. Then we take the positive root
[tex]sin(t) =\sqrt{\frac{7}{16}}[/tex]
[tex]sin(t) =\frac{\sqrt{7}}{4}[/tex]