Answer:
[tex]sec(2\theta)=-\frac{5}{4}[/tex]
Step-by-step explanation:
we know that
step 1
Find [tex]cos(\theta)[/tex]
we know that
[tex]tan^2(\theta)+1=sec^2(\theta)[/tex]
we have
[tex]tan(\theta)=3[/tex]
substitute
[tex]3^2+1=sec^2(\theta)[/tex]
[tex]sec^2(\theta)=10[/tex]
[tex]sec(\theta)=\pm\sqrt{10}[/tex]
Remember that
Angle theta lie in Quadrant I
so
[tex]sec(\theta)[/tex] is positive
[tex]sec(\theta)=\sqrt{10}[/tex]
Remember that
[tex]sec(\theta)=\frac{1}{cos(\theta)}[/tex]
therefore
[tex]cos(\theta)=\frac{1}{\sqrt{10}}[/tex]
step 2
Find [tex]sin(\theta)[/tex]
we know that
[tex]tan(\theta)=\frac{sin(\theta)}{cos(\theta)}[/tex]
[tex]sin(\theta)=tan(\theta)cos(\theta)[/tex]
we have
[tex]cos(\theta)=\frac{1}{\sqrt{10}}[/tex]
[tex]tan(\theta)=3[/tex]
substitute
[tex]sin(\theta)=(3)(\frac{1}{\sqrt{10}})[/tex]
[tex]sin(\theta)=\frac{3}{\sqrt{10}}[/tex]
step 3
Find [tex]cos(2\theta)[/tex]
we know that
[tex]cos(2\theta)=2cos^2(\theta)-1[/tex]
we have
[tex]cos(\theta)=\frac{1}{\sqrt{10}}[/tex]
substitute
[tex]cos(2\theta)=2(\frac{1}{\sqrt{10}})^2-1[/tex]
[tex]cos(2\theta)=\frac{1}{5}-1[/tex]
[tex]cos(2\theta)=-\frac{4}{5}[/tex]
Remember that
[tex]sec(2\theta)=\frac{1}{cos(2\theta)}[/tex]
therefore
[tex]sec(2\theta)=-\frac{5}{4}[/tex]
