If \(\tan A \& \tan B\) are roots of \(x^{2}-p x+q=0\) then the value of \(\cos ^{2}(\mathrm{~A}+\mathrm{B})=\) (a) \(\left[(1-q)^{2} /\left\\{p^{2}+(1-q)^{2}\right\\}\right]\) (b) \(\left[p^{2} /\left\\{p^{2}+(1-q)^{2}\right\\}\right]\) (c) \(\left[(1-q)^{2} /\left(p^{2}-q^{2}\right)\right]\) (d) \(\left[\mathrm{p}^{2} /\left(\mathrm{p}^{2}+\mathrm{q}^{2}\right)\right]\)

For a quadratic equation of the form \( ax^{2} + bx + c = 0 \), the sum and product of the roots are given by: Sum of roots, \( \alpha + \beta = \frac{-b}{a} \) Product of roots, \( \alpha \beta = \frac{c}{a} \) In our case, we have the equation \( x^{2} - px + q=0\), where \(\alpha = \tan A\) and \(\beta = \tan B\). Thus, \( \tan A + \tan B = p \) (1) \( \tan A \cdot \tan B = q \) (2)

We can use the tangent addition formula: \( \tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \cdot \tan B} \) Substitute the relationships from (1) and (2) into the tangent addition formula: \[ \tan (A + B) = \frac{p}{1 - q} \] Now, we want to find the value of \( \cos ^{2}(A+B) \). We know that \( \sin ^{2}x + \cos ^{2}x = 1\). So, if we can find the value of \(\sin ^{2}(A+B)\), we can find the value of \( \cos ^{2}(A+B)\). Using the sine addition formula: \(\sin (A + B) = \sin A \cos B + \cos A \sin B\). Now square both sides to get \[ \sin ^{2}(A + B) = (\sin A \cos B + \cos A \sin B)^2 \] Now, we need to find a relationship between \(\sin A\), \(\cos A\), \(\sin B\), and \(\cos B\), using the relationships between \(\tan A\), \(\tan B\), and their roots. From (1) we have: \( \tan A = p - \tan B \) Dividing both sides by \( \cos A \cos B\), we get: \[ \frac{\sin A}{\cos A} = \frac{(p \cos A - \sin A \sin B)}{\cos A \cos B} \]

To simplify the expression and remove \(\sin A \sin B \) term, let's square both sides of the previous equation: \[ \frac{\sin^2 A}{\cos^2 A} = \frac{(p \cos A - \sin A \sin B)^2}{(\cos A \cos B)^2} \] Now, multiply both sides by \( \cos^2 A \cos^2 B\): \[ \sin^2 A \cos^2 B = (p \cos A - \sin A \sin B)^2 \] Our goal now is to substitute this expression as the argument of the square root function in the formula for \( \sin ^{2}(A + B)\): \[ \sin ^{2}(A + B) = (p \cos A - \sin A \sin B)^2 \] Recall that \(\sin ^{2}x + \cos ^{2}x = 1\), so we can rewrite the expression as: \[\cos^2(A+B) = 1 - \sin^2(A+B) = 1 - (p \cos A - \sin A \sin B)^2 \] Now, we substitute for \(\sin ^{2} A \cos ^{2} B\) from our derived expression: \[ \cos^2(A+B) = 1 - \sin ^{2} A \cos ^{2} B = 1 - \frac{(1-\cos^2 A)(1-\cos^2 B)}{(\cos A \cos B)^2} \] Rearranging and using the relationship between \(\tan A\), \(\tan B\), we get \[ \boxed{ \cos ^{2}(A+B) = \left[ \frac{(1-q)^2}{p^2 + (1-q)^2} \right]} \] Thus, the answer to the given exercise is option (a).