Key Concepts and Formulas

• Vieta's Formulas: For a quadratic equation $ax^2 + bx + c = 0$, the sum of the roots is $-b/a$ and the product of the roots is $c/a$.
• Tangent Addition Formula: $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$.
• Trigonometric Identity: $\sec^2\theta = 1 + \tan^2\theta$, and $\cos^2\theta = \frac{1}{\sec^2\theta} = \frac{1}{1 + \tan^2\theta}$.

Step-by-Step Solution

1. Applying Vieta's Formulas The given quadratic equation is $3x^2 - 10x - 25 = 0$. Since $\tan A$ and $\tan B$ are the roots, we can use Vieta's formulas to find the sum and product of the roots.

• Sum of roots: $\tan A + \tan B = -(-10)/3 = \frac{10}{3}$.
• Product of roots: $\tan A \tan B = -25/3$.

Why this step? This step extracts the fundamental relationships between $\tan A$ and $\tan B$ from the given equation, which are crucial for calculating $\tan(A+B)$.

2. Calculating $\tan(A+B)$ Now, we use the tangent addition formula: $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$ Substitute the values from Step 1: $\tan(A + B) = \frac{\frac{10}{3}}{1 - (-\frac{25}{3})} = \frac{\frac{10}{3}}{1 + \frac{25}{3}} = \frac{\frac{10}{3}}{\frac{28}{3}} = \frac{10}{28} = \frac{5}{14}$

Why this step? We need to evaluate an expression involving $(A+B)$. Knowing $\tan(A+B)$ is the first step towards evaluating trigonometric functions of $(A+B)$.

3. Rewriting the Target Expression Let the expression to be evaluated be $E$. $E = 3 \sin^2(A + B) - 10 \sin(A + B)\cos(A + B) - 25 \cos^2(A + B)$ Let $\theta = A + B$. The expression becomes: $E = 3 \sin^2\theta - 10 \sin\theta \cos\theta - 25 \cos^2\theta$ To relate this to $\tan\theta$, we can divide each term by $\cos^2\theta$ (assuming $\cos\theta \neq 0$). This gives us: $\frac{E}{\cos^2\theta} = \frac{3 \sin^2\theta}{\cos^2\theta} - \frac{10 \sin\theta \cos\theta}{\cos^2\theta} - \frac{25 \cos^2\theta}{\cos^2\theta}$ $\frac{E}{\cos^2\theta} = 3 \tan^2\theta - 10 \tan\theta - 25$ Therefore, we can write $E$ as: $E = \cos^2\theta (3 \tan^2\theta - 10 \tan\theta - 25)$

Why this step? This transformation allows us to use the previously calculated value of $\tan(A+B)$. It also reveals a connection to the original quadratic polynomial.

4. Evaluating the Polynomial Term The term in the parenthesis, $3 \tan^2\theta - 10 \tan\theta - 25$, is the polynomial $P(x) = 3x^2 - 10x - 25$ evaluated at $x = \tan\theta = \tan(A+B)$. Let $t = \tan(A+B) = \frac{5}{14}$. We need to calculate $3t^2 - 10t - 25$: $3\left(\frac{5}{14}\right)^2 - 10\left(\frac{5}{14}\right) - 25 = 3\left(\frac{25}{196}\right) - \frac{50}{14} - 25 = \frac{75}{196} - \frac{700}{196} - \frac{4900}{196} = \frac{75 - 700 - 4900}{196} = \frac{-5525}{196}$

Why this step? This calculates the value of the quadratic part of the expression with $\tan(A+B)$.

5. Calculating $\cos^2(A+B)$ We use the identity $\cos^2\theta = \frac{1}{1 + \tan^2\theta}$. With $\tan(A+B) = \frac{5}{14}$: $\tan^2(A+B) = \left(\frac{5}{14}\right)^2 = \frac{25}{196}$ $\cos^2(A+B) = \frac{1}{1 + \frac{25}{196}} = \frac{1}{\frac{196 + 25}{196}} = \frac{1}{\frac{221}{196}} = \frac{196}{221}$

Why this step? This provides the other necessary component, $\cos^2(A+B)$, to complete the evaluation of $E$.

6. Final Calculation of E Now substitute the results from Steps 4 and 5 back into the rewritten expression for $E$: $E = \cos^2(A+B) (3 \tan^2(A+B) - 10 \tan(A+B) - 25)$ $E = \left(\frac{196}{221}\right) \times \left(\frac{-5525}{196}\right) = \frac{-5525}{221} = -25$

Why this step? This combines all the calculated parts to arrive at the final numerical value of the expression.

Common Mistakes & Tips

• Sign Errors: Be extremely careful with negative signs, especially when applying Vieta's formulas and the tangent addition formula.
• Division by Zero: Remember that dividing by $\cos^2(A+B)$ is only valid if $\cos(A+B) \neq 0$. However, if $\cos(A+B) = 0$, then $\tan(A+B)$ is undefined, which contradicts the fact that $\tan(A+B) = 5/14$. Therefore, we are safe.
• Trigonometric Identities: Double-check the trigonometric identities you are using to ensure they are correct.

Summary

The problem elegantly combines quadratic equations and trigonometry. By using Vieta's formulas, we find $\tan A + \tan B$ and $\tan A \tan B$, which allows us to calculate $\tan(A+B)$. Rewriting the target expression in terms of $\tan(A+B)$ allows us to exploit the connection to the original quadratic equation. The final calculation involves combining the value of the quadratic expression evaluated at $\tan(A+B)$ with the value of $\cos^2(A+B)$. The value of the given expression is $-25$.

Final Answer

The final answer is $\boxed{-25}$, which corresponds to option (C).