Question

HELP FAST PLS QUICK!!!!

Answer 1

Answer:

$\sqrt{2}$

Step-by-step explanation:

we know that

$tan^{2}(\theta)+1=sec^{2}(\theta)$

we have

$tan^{2}(\theta)=-1$

substitute the value

$(-1^{2})+1=sec^{2}(\theta)$

$(1)+1=sec^{2}(\theta)$

$sec^{2}(\theta)=2\\ \\sec(\theta)=(+/-)\sqrt{2}$

Remember that

The angle theta belong to the III and IV quadrant, but the tangent is negative, therefore belong to the IV quadrant

so

$sec(\theta)$ is positive

therefore

$sec(\theta)=+\sqrt{2}$