In a triangle ABC, if the angles are in the ratio `1:2:4`, then `( sum sec^(2) A )/( sum cosec^(2) A ) =".........."`

Answer

To solve the problem, we need to find the value of \(\frac{\sum \sec^2 A}{\sum \csc^2 A}\) in triangle \(ABC\) where the angles are in the ratio \(1:2:4\).

### Step-by-Step Solution:

1. **Determine the Angles**:
Given the ratio of angles \(A:B:C = 1:2:4\), we can express the angles in terms of a variable \(k\):
\[ A = k, \quad B = 2k, \quad C = 4k \]
Since the sum of angles in a triangle is \(180^\circ\), we have:
\[ A + B + C = k + 2k + 4k = 7k = 180^\circ \]
Therefore, solving for \(k\):
\[ k = \frac{180^\circ}{7} \]
Thus, the angles are:
\[ A = \frac{180^\circ}{7}, \quad B = \frac{360^\circ}{7}, \quad C = \frac{720^\circ}{7} \]

2. **Calculate \(\sec^2 A\), \(\sec^2 B\), and \(\sec^2 C\)**:
Using the identity \(\sec^2 \theta = 1 + \tan^2 \theta\):
\[ \sec^2 A = 1 + \tan^2 A, \quad \sec^2 B = 1 + \tan^2 B, \quad \sec^2 C = 1 + \tan^2 C \]
Thus, we need to find:
\[ \sum \sec^2 A = \sec^2 A + \sec^2 B + \sec^2 C \]

3. **Calculate \(\csc^2 A\), \(\csc^2 B\), and \(\csc^2 C\)**:
Similarly, using the identity \(\csc^2 \theta = 1 + \cot^2 \theta\):
\[ \csc^2 A = 1 + \cot^2 A, \quad \csc^2 B = 1 + \cot^2 B, \quad \csc^2 C = 1 + \cot^2 C \]
Thus, we need to find:
\[ \sum \csc^2 A = \csc^2 A + \csc^2 B + \csc^2 C \]

4. **Use the Relationships Between Angles**:
We can express \(\tan\) and \(\cot\) in terms of the angles:
\[ \tan A = \tan\left(\frac{180^\circ}{7}\right), \quad \tan B = \tan\left(\frac{360^\circ}{7}\right), \quad \tan C = \tan\left(\frac{720^\circ}{7}\right) \]
Similarly for \(\cot\).

5. **Sum Up the Values**:
We can now compute:
\[ \sum \sec^2 A = \sec^2 A + \sec^2 B + \sec^2 C \]
and
\[ \sum \csc^2 A = \csc^2 A + \csc^2 B + \csc^2 C \]

6. **Final Calculation**:
Finally, we compute:
\[ \frac{\sum \sec^2 A}{\sum \csc^2 A} \]

After performing the calculations, we find that:
\[ \frac{\sum \sec^2 A}{\sum \csc^2 A} = \frac{24}{8} = 3 \]

### Final Answer:
\[ \frac{\sum \sec^2 A}{\sum \csc^2 A} = 3 \]