If [tex]\theta[/tex] is in the first quadrant, we know that [tex]\cos\theta[/tex] is positive.
Recall the Pythagorean identity,
[tex]\cos^2\theta+\sin^2\theta=1[/tex]
Dividing both sides by [tex]\cos^2\theta[/tex] gives
[tex]\dfrac{\cos^2\theta}{\cos^2\theta}+\dfrac{\sin^2\theta}{\cos^2\theta}=\dfrac1{\cos^2\theta}\iff1+\tan^2\theta=\sec^2\theta[/tex]
Because
[tex]\sec\theta=\dfrac1{\cos\theta}[/tex]
if [tex]\cos\theta>0[/tex], then we also have [tex]\sec\theta>0[/tex]. This means that we take the square root of both sides in the [tex]\tan[/tex]-[tex]\sec[/tex] identity, we take the positive square root and we get
[tex]\sec\theta=\sqrt{1+\left(\dfrac{40}9\right)^2}=\dfrac{41}9[/tex]
[tex]\implies\cos\theta=\dfrac9{41}[/tex]
Now, recall the double angle identity,
[tex]\cos2\theta=\cos^2\theta-\sin^2\theta[/tex]
Apply the Pythagorean identity to eliminate [tex]\sin\theta[/tex]:
[tex]\cos2\theta=\cos^2\theta-(1-\cos^2\theta)=2\cos^2\theta-1[/tex]
So we end up with
[tex]\cos2\theta=2\left(\dfrac9{41}\right)^2-1=-\dfrac{1519}{1681}[/tex]
