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Step-by-Step Solution
Step 1: Identify the sum and product of the roots
Given the quadratic equation
$$3x^2 - 10x - 25 = 0,$$
let its roots be
$$\tan A \quad \text{and} \quad \tan B.$$
By Viète's formulas:
Sum of roots,
$$\tan A + \tan B = \frac{10}{3}.$$
Product of roots,
$$\tan A \cdot \tan B = \frac{-25}{3}.$$
Step 2: Compute $ \tan(A + B) $
Recall the identity for the tangent of a sum:
$$\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \,\tan B}.$$
Substitute the sum and product of the roots:
$$\tan(A + B)
= \frac{\frac{10}{3}}{1 - \left(\frac{-25}{3}\right)}
= \frac{\frac{10}{3}}{1 + \frac{25}{3}}
= \frac{\frac{10}{3}}{\frac{3 + 25}{3}}
= \frac{10}{3} \times \frac{3}{28}
= \frac{5}{14}.$$
Step 3: Rewrite the expression in terms of $ \tan(A + B) $
We need to evaluate:
$$3\,\sin^2(A + B)\;-\;10\,\sin(A + B)\,\cos(A + B)\;-\;25\,\cos^2(A + B).$$
Use the following identities involving $ \theta = A + B $:
$\sin^2 \theta = \frac{\tan^2 \theta}{1 + \tan^2 \theta},$
$\cos^2 \theta = \frac{1}{1 + \tan^2 \theta},$
$\sin \theta \cos \theta = \frac{\tan \theta}{1 + \tan^2 \theta}.$
Thus,
\[
\begin{aligned}
3\,\sin^2 \theta &- 10\,\sin \theta \cos \theta \;-\; 25\,\cos^2 \theta \\
&= 3 \cdot \frac{\tan^2 \theta}{1 + \tan^2 \theta} \;-\; 10 \cdot \frac{\tan \theta}{1 + \tan^2 \theta} \;-\; 25 \cdot \frac{1}{1 + \tan^2 \theta}.
\end{aligned}
\]
Factor out $\frac{1}{1 + \tan^2 \theta}$:
\[
= \frac{1}{1 + \tan^2 \theta}\;\bigl(3\,\tan^2 \theta - 10\,\tan \theta - 25\bigr).
\]
Step 4: Substitute $ \tan(A + B) = \frac{5}{14} $ and simplify
Let
$$\theta = A + B,\quad \tan \theta = \frac{5}{14}.$$
1) Compute $1 + \tan^2 \theta$:
\[
\tan^2 \theta = \left(\frac{5}{14}\right)^2 = \frac{25}{196},
\]
\[
1 + \tan^2 \theta = 1 + \frac{25}{196} = \frac{196 + 25}{196} = \frac{221}{196}.
\]
Hence,
$$
\frac{1}{1 + \tan^2 \theta} = \frac{196}{221}.
$$
2) Evaluate the bracket $\bigl(3\,\tan^2 \theta - 10\,\tan \theta - 25\bigr)$:
\[
3\,\tan^2 \theta = 3 \times \frac{25}{196} = \frac{75}{196},
\]
\[
-\,10\,\tan \theta = -\,10 \times \frac{5}{14} = -\,\frac{50}{14} = -\,\frac{25}{7}
= -\,\frac{25 \times 28}{7 \times 28} = -\,\frac{700}{196},
\]
\[
-\,25 = -\,\frac{25 \times 196}{196} = -\,\frac{4900}{196}.
\]
Adding them together:
\[
\frac{75}{196} - \frac{700}{196} - \frac{4900}{196}
= \frac{75 - 700 - 4900}{196}
= \frac{-5525}{196}.
\]
3) Multiply by $ \frac{196}{221} $:
\[
\left(- \frac{5525}{196}\right) \times \frac{196}{221}
= - \frac{5525}{221}.
\]
Notice $5525 = 25 \times 221,$ so:
\[
- \frac{5525}{221} = - 25.
\]
Therefore, the value of
$$3\,\sin^2(A + B) \;-\; 10\,\sin(A + B)\,\cos(A + B) \;-\; 25\,\cos^2(A + B)$$
is
$$\boxed{-25}.$$
