Recall that
[tex]1+\tan^2x=\sec^2x\implies\sec x=\sqrt{1+\tan^2x}[/tex]
We know to take the positive root here because [tex]\cos x>0[/tex], which means [tex]\sec x>0[/tex] as well. Then
[tex]\sec x=\sqrt{1+3^2}=\sqrt4=2\implies\cos x=\dfrac12[/tex]
We then have
[tex]\tan x=\dfrac{\sin x}{\cos x}=\dfrac{\sin x}{\frac12}=3\implies\sin x=\dfrac32\implies\sec x=\dfrac23[/tex]
