If $\sqrt{9−8\cos 40^{\circ}} = a +b\sec 40^{\circ}$, then what is $|a+b|$?

Question

Suppose $\sqrt{9−8\cos 40^{\circ}} = a +b\sec 40^{\circ}$, where $a$ and $b$ are rational numbers.

Then we can say $|a + b|$ equals $2$?

I tried but didn't found way to approach this problem. I tried taking the value of $\cos 40^{\circ},$ and then $\sqrt{9-8\cos 40^{\circ}} \approx 1.7$ and $\sec 40^{\circ} \approx 1.3$.

Then took $a=0.4$ and $b=1$ to satisfy this case. So $|a+b|$ comes to $1.4$, but the correct answer is $2$. Can't figure whats wrong.

Answer 1

$\cos(3\cdot40^{\circ})=-\frac{1}{2}$ gives $$8\cos^340^{\circ}-6\cos40^{\circ}+1=0.$$

In another hand, $$9\cos^240^{\circ}-8\cos^340^{\circ}=(a\cos40^{\circ}+b)^2$$ or $$9\cos^240^{\circ}-(6\cos40^{\circ}-1)=(a\cos40^{\circ}+b)^2,$$ which gives $a^2=9$, $b^2=1$ and $2ab=-6.$

Can you end it now?