If tanA and tanB the roots of the quadratic equation, `4x^(2)-7x+1=0` then evaluate `4sin^(2)(A+B)-7sin(A+B)*cos(A+B)+cos^(2)(A+B)`.

Answer

To solve the problem, we need to evaluate the expression \(4\sin^2(A+B) - 7\sin(A+B)\cos(A+B) + \cos^2(A+B)\) given that \(\tan A\) and \(\tan B\) are the roots of the quadratic equation \(4x^2 - 7x + 1 = 0\).

### Step 1: Find the values of \(\tan A + \tan B\) and \(\tan A \tan B\)
Using Vieta's formulas, for the quadratic equation \(ax^2 + bx + c = 0\):
- The sum of the roots (\(\tan A + \tan B\)) is given by \(-\frac{b}{a}\).
- The product of the roots (\(\tan A \tan B\)) is given by \(\frac{c}{a}\).

For our equation \(4x^2 - 7x + 1 = 0\):
- \(\tan A + \tan B = \frac{7}{4}\)
- \(\tan A \tan B = \frac{1}{4}\)

### Step 2: Find \(\tan(A+B)\)
Using the formula for the tangent of the sum of two angles:
\[ \tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]
Substituting the values we found:
\[ \tan(A+B) = \frac{\frac{7}{4}}{1 - \frac{1}{4}} = \frac{\frac{7}{4}}{\frac{3}{4}} = \frac{7}{3} \]

### Step 3: Calculate \(\sin(A+B)\) and \(\cos(A+B)\)
Using the identity:
\[ \sin^2(A+B) + \cos^2(A+B) = 1 \]
And the relation:
\[ \tan(A+B) = \frac{\sin(A+B)}{\cos(A+B)} \]
Let \(\sin(A+B) = k\) and \(\cos(A+B) = \sqrt{1 - k^2}\). Then:
\[ \frac{k}{\sqrt{1-k^2}} = \frac{7}{3} \]
Squaring both sides:
\[ k^2 = \frac{49}{58} \]
Thus, \(\sin^2(A+B) = \frac{49}{58}\) and \(\cos^2(A+B) = 1 - \sin^2(A+B) = \frac{9}{58}\).

### Step 4: Substitute into the expression
Now we substitute \(\sin^2(A+B)\) and \(\cos^2(A+B)\) into the expression:
\[ 4\sin^2(A+B) - 7\sin(A+B)\cos(A+B) + \cos^2(A+B) \]
Calculating each term:
1. \(4\sin^2(A+B) = 4 \cdot \frac{49}{58} = \frac{196}{58}\)
2. \(-7\sin(A+B)\cos(A+B) = -7 \cdot \sqrt{\frac{49}{58}} \cdot \sqrt{\frac{9}{58}} = -7 \cdot \frac{21}{58} = -\frac{147}{58}\)
3. \(\cos^2(A+B) = \frac{9}{58}\)

Putting it all together:
\[ \frac{196}{58} - \frac{147}{58} + \frac{9}{58} = \frac{196 - 147 + 9}{58} = \frac{58}{58} = 1 \]

### Final Answer
Thus, the value of the expression is:
\[ \boxed{1} \]