### Instruction: If $A$ is an angle such that $\tan A + \sec A = 2,$ enter all possible values of $\cos A,$ separated by commas.### Response: I know that $\tan A = \frac{\sin A}{\cos A}$ and $\sec A = \frac{1}{\cos A},$ so I can rewrite the given equation as $\frac{\sin A}{\cos A} + \frac{1}{\cos A} = 2.$
To simplify this equation, I can multiply both sides by $\cos A,$ assuming $\cos A \neq 0,$ to get $\sin A + 1 = 2 \cos A.$
Now I can rearrange the equation to get $2 \cos A - \sin A = 1.$
This looks like a Pythagorean identity, so I can try to use $\cos^2 A + \sin^2 A = 1$ to eliminate one of the trigonometric functions.
If I square both sides of the equation, I get $4 \cos^2 A - 4 \cos A \sin A + \sin^2 A = 1.$
Then I can substitute $1 - \cos^2 A$ for $\sin^2 A$ using the identity, and simplify to get $3 \cos^2 A - 4 \cos A \sin A = 0.$
I can factor out $\cos A$ from this equation, and again assume $\cos A \neq 0,$ to get $\cos A (3 \cos A - 4 \sin A) = 0.$
This means either $\cos A = 0$ or $3 \cos A - 4 \sin A = 0.$
If $\cos A = 0,$ then $\sin A$ must be $\pm 1,$ but then the original equation would not hold, since $\tan A$ and $\sec A$ would be undefined.
So I can rule out this possibility.
If $3 \cos A - 4 \sin A = 0,$ then I can divide both sides by $\cos A,$ again assuming $\cos A \neq 0,$ to get $3 - 4 \tan A = 0.$
This implies that $\tan A = \frac{3}{4},$ which means that $\cos A = \frac{4}{5}$ or $\cos A = -\frac{4}{5},$ depending on the quadrant of $A.$
I can check that both values of $\cos A$ satisfy the original equation by plugging them in and verifying that $\tan A + \sec A = 2.$
Therefore, the possible values of $\cos A$ are $\frac{4}{5}$ and $-\frac{4}{5}.$