To find the sum `sin^(2) ""(2pi)/(7) + sin^(2)""(4pi)/(7) +sin^(2)""(8pi)/(7)`, we follow the following method.
Put `7theta = 2npi`, where `n ` is any integer. Then
`" " sin 4 theta = sin( 2npi - 3theta) = - sin 3theta`
This means that `sin theta` takes the values `0, pm sin (2pi//7), pmsin(2pi//7), pm sin(4pi//7), and pm sin (8pi//7)`.
From Eq. (i), we now get
`" " 2 sin 2 theta cos 2theta = 4 sin^(3) theta - 3 sin theta `
or `4 sin theta cos theta (1-2 sin^(2) theta)= sin theta ( 4sin ^(2) theta -3)`
Rejecting the value `sin theta =0`, we get
`" " 4 cos theta (1-2 sin^(2) theta ) = 4 sin ^(2) theta - 3`
or ` 16 cos^(2) theta (1-2 sin^(2) theta)^(2) = ( 4sin ^(2) theta -3)^(2)`
or `16(1-sin^(2) theta) (1-4 sin^(2) theta + 4 sin ^(4) theta)`
`" " = 16 sin ^(4) theta - 24 sin ^(2) theta +9`
or `" " 64 sin^(6) theta - 112 sin^(4) theta - 56 sin^(2) theta -7 =0`
This is cubic in `sin^(2) theta` with the roots `sin^(2)( 2pi//7), sin^(2) (4pi//7), and sin^(2)(8pi//7)`.
The sum of these roots is
`" " sin^(2)""(2pi)/(7) + sin^(2)""(4pi)/(7) + sin ^(2)""(8pi)/(7) = (112)/(64) = (7)/(4)`.
The value of `(tan^(2)""(pi)/(7) + tan^(2)""(2pi)/(7) + tan^(2)""(3pi)/(7))xx (cot^(2)""(pi)/(7) + cot^(2)""(2pi)/(7) + cot^(2)""(3pi)/(7))` is

Options

A. 105
B. 35
C. 210
D. none of these.

Explanation

To solve the problem, we need to find the value of the expression:

\[ \left( \tan^2 \frac{\pi}{7} + \tan^2 \frac{2\pi}{7} + \tan^2 \frac{3\pi}{7} \right) \left( \cot^2 \frac{\pi}{7} + \cot^2 \frac{2\pi}{7} + \cot^2 \frac{3\pi}{7} \right) \]

### Step 1: Define the variables
Let:
\[ z = \tan \theta \quad \text{where } \theta = \frac{\pi}{7}, \frac{2\pi}{7}, \frac{3\pi}{7} \]

### Step 2: Use the identity for tangent
Using the identity for tangent:
\[ \tan(4\theta) = \frac{4\tan \theta - 4\tan^3 \theta}{1 - 6\tan^2 \theta + \tan^4 \theta} \]
and
\[ \tan(3\theta) = \frac{3\tan \theta - \tan^3 \theta}{1 - 3\tan^2 \theta} \]

### Step 3: Set up the equation
From the above identities, we can set up the equation:
\[ \frac{4z - 4z^3}{1 - 6z^2 + z^4} = -\frac{3z - z^3}{1 - 3z^2} \]

### Step 4: Cross-multiply
Cross-multiplying gives:
\[ (4z - 4z^3)(1 - 3z^2) = -(3z - z^3)(1 - 6z^2 + z^4) \]

### Step 5: Simplify the equation
Expanding both sides:
\[ 4z - 12z^3 - 4z^3 + 12z^5 = -3z + 18z^3 - z^5 + 6z^5 - 3z^7 \]
Combine like terms to form a polynomial equation.

### Step 6: Rearranging
Rearranging gives us a cubic equation in \(z^2\):
\[ z^6 - 21z^4 + 35z^2 - 7 = 0 \]

### Step 7: Roots of the cubic equation
The roots of this cubic equation correspond to \(\tan^2 \frac{\pi}{7}\), \(\tan^2 \frac{2\pi}{7}\), and \(\tan^2 \frac{3\pi}{7}\).

### Step 8: Sum of the roots
Using the property of the sum of the roots for a cubic equation \(ax^3 + bx^2 + cx + d = 0\):
\[ \text{Sum of roots} = -\frac{b}{a} = 21 \]

### Step 9: Finding cotangent values
Since \(\cot^2 \theta = \frac{1}{\tan^2 \theta}\), we can find the sum of the cotangent squares:
\[ \cot^2 \frac{\pi}{7} + \cot^2 \frac{2\pi}{7} + \cot^2 \frac{3\pi}{7} = \frac{1}{\tan^2 \frac{\pi}{7}} + \frac{1}{\tan^2 \frac{2\pi}{7}} + \frac{1}{\tan^2 \frac{3\pi}{7}} \]
This can be expressed as:
\[ \frac{\tan^2 \frac{2\pi}{7} \tan^2 \frac{3\pi}{7} + \tan^2 \frac{\pi}{7} \tan^2 \frac{3\pi}{7} + \tan^2 \frac{\pi}{7} \tan^2 \frac{2\pi}{7}}{\tan^2 \frac{\pi}{7} \tan^2 \frac{2\pi}{7} \tan^2 \frac{3\pi}{7}} \]

### Step 10: Sum of cotangent roots
Using the property of roots for the reciprocal polynomial, we find:
\[ \text{Sum of roots} = 5 \]

### Step 11: Final calculation
Now we can calculate the final expression:
\[ \left( \tan^2 \frac{\pi}{7} + \tan^2 \frac{2\pi}{7} + \tan^2 \frac{3\pi}{7} \right) \left( \cot^2 \frac{\pi}{7} + \cot^2 \frac{2\pi}{7} + \cot^2 \frac{3\pi}{7} \right) = 21 \times 5 = 105 \]

### Conclusion
Thus, the value of the expression is:
\[ \boxed{105} \]