If in `DeltaABC, tan A+tanB+tanC=6` and `tanA tanB=2`, then `sin^(2)A:sin^(2)B:sin^(2)C` is

Options

A. `8:9:5`
B. `8:5:9`
C. `5:9:5`
D. `5:8:5`

Answer

To solve the problem, we will use the given information about the triangle \( \Delta ABC \) and the properties of trigonometric functions.

### Step 1: Use the given information
We are given:
1. \( \tan A + \tan B + \tan C = 6 \)
2. \( \tan A \tan B = 2 \)

Since \( A + B + C = 180^\circ \) in a triangle, we can use the identity:
\[ \tan A + \tan B + \tan C = \tan A \tan B \tan C \]
This means:
\[ 6 = 2 \tan C \]
From this, we can solve for \( \tan C \):
\[ \tan C = \frac{6}{2} = 3 \]

### Step 2: Find \( \tan A \) and \( \tan B \)
Now we know \( \tan C = 3 \). We can substitute this back into the equation for \( \tan A + \tan B \):
\[ \tan A + \tan B = 6 - \tan C = 6 - 3 = 3 \]

Now we have two equations:
1. \( \tan A + \tan B = 3 \)
2. \( \tan A \tan B = 2 \)

Let \( \tan A = x \) and \( \tan B = y \). Thus, we have:
\[ x + y = 3 \]
\[ xy = 2 \]

### Step 3: Form a quadratic equation
We can form a quadratic equation using these two equations:
\[ t^2 - (x+y)t + xy = 0 \]
Substituting the values:
\[ t^2 - 3t + 2 = 0 \]

### Step 4: Solve the quadratic equation
Now we can solve this quadratic equation using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -3, c = 2 \):
\[ t = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1} = \frac{3 \pm \sqrt{9 - 8}}{2} = \frac{3 \pm 1}{2} \]
This gives us:
\[ t = \frac{4}{2} = 2 \quad \text{or} \quad t = \frac{2}{2} = 1 \]
Thus, \( \tan A = 2 \) and \( \tan B = 1 \) (or vice versa).

### Step 5: Find \( \sin^2 A, \sin^2 B, \sin^2 C \)
Now we can find \( \sin^2 A, \sin^2 B, \sin^2 C \) using the relationship:
\[ \sin^2 \theta = \frac{\tan^2 \theta}{1 + \tan^2 \theta} \]

Calculating for \( A \):
\[ \sin^2 A = \frac{\tan^2 A}{1 + \tan^2 A} = \frac{2^2}{1 + 2^2} = \frac{4}{5} \]

Calculating for \( B \):
\[ \sin^2 B = \frac{\tan^2 B}{1 + \tan^2 B} = \frac{1^2}{1 + 1^2} = \frac{1}{2} \]

Calculating for \( C \):
\[ \sin^2 C = \frac{\tan^2 C}{1 + \tan^2 C} = \frac{3^2}{1 + 3^2} = \frac{9}{10} \]

### Step 6: Find the ratio \( \sin^2 A : \sin^2 B : \sin^2 C \)
We have:
- \( \sin^2 A = \frac{4}{5} \)
- \( \sin^2 B = \frac{1}{2} \)
- \( \sin^2 C = \frac{9}{10} \)

To find the ratio, we can express them with a common denominator. The least common multiple of \( 5, 2, \) and \( 10 \) is \( 10 \):
- \( \sin^2 A = \frac{4}{5} = \frac{8}{10} \)
- \( \sin^2 B = \frac{1}{2} = \frac{5}{10} \)
- \( \sin^2 C = \frac{9}{10} \)

Thus, the ratio is:
\[ \sin^2 A : \sin^2 B : \sin^2 C = 8 : 5 : 9 \]

### Final Answer
The ratio \( \sin^2 A : \sin^2 B : \sin^2 C \) is \( 8 : 5 : 9 \).