Answer:
[tex]\sqrt{2}[/tex]
Step-by-step explanation:
we know that
[tex]tan^{2}(\theta)+1=sec^{2}(\theta)[/tex]
we have
[tex]tan^{2}(\theta)=-1[/tex]
substitute the value
[tex](-1^{2})+1=sec^{2}(\theta)[/tex]
[tex](1)+1=sec^{2}(\theta)[/tex]
[tex]sec^{2}(\theta)=2\\ \\sec(\theta)=(+/-)\sqrt{2}[/tex]
Remember that
The angle theta belong to the III and IV quadrant, but the tangent is negative, therefore belong to the IV quadrant
so
[tex]sec(\theta)[/tex] is positive
therefore
[tex]sec(\theta)=+\sqrt{2}[/tex]
