In `triangleABC`, which is not right angled, if p=sinA sinB sinC = and q=cosA. cosB. cosC. Then the equation having roots tanA,tanB and tanC is

Options

A. `qx^(3) - px^(2) + (1+q)x-p=0`
B. `qx^(3)+2px^(2)+qx-p=0`
C. `qx^(3)-px^(2)+(2+q)x+pq=0`
D. `qx^(3)-px^(2)+qx+p=0`

Answer

To solve the problem, we need to find the cubic equation whose roots are \( \tan A, \tan B, \tan C \) given the conditions \( p = \sin A \sin B \sin C \) and \( q = \cos A \cos B \cos C \).

### Step-by-Step Solution:

1. **Understanding the Roots**:
We know that the roots of the cubic equation are \( \tan A, \tan B, \tan C \).

2. **Finding the Product of Roots**:
The product of the roots \( \tan A \tan B \tan C \) can be expressed using the given values of \( p \) and \( q \):
\[ \tan A \tan B \tan C = \frac{p}{q} \]

3. **Finding the Sum of Roots**:
In any triangle, the sum of angles is \( A + B + C = 180^\circ \) or \( \pi \) radians. Therefore, we can use the identity:
\[ \tan(A + B) = \tan C \]
This leads to:
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} = \tan C \]
Rearranging gives:
\[ \tan A + \tan B + \tan C = \tan A \tan B \tan C \]
Substituting the product of roots:
\[ \tan A + \tan B + \tan C = \frac{p}{q} \]

4. **Finding the Sum of Product of Roots**:
We need to find \( \tan A \tan B + \tan B \tan C + \tan C \tan A \). Using the identities:
\[ \tan A \tan B = \frac{1 + \cos C}{\cos A \cos B} \]
\[ \tan B \tan C = \frac{1 + \cos A}{\cos B \cos C} \]
\[ \tan C \tan A = \frac{1 + \cos B}{\cos C \cos A} \]
Adding these gives:
\[ \tan A \tan B + \tan B \tan C + \tan C \tan A = \frac{3 + \left( \cos A + \cos B + \cos C \right)}{\cos A \cos B \cos C} \]
Using the identity \( \cos^2 A + \cos^2 B + \cos^2 C = 1 - 2 \cos A \cos B \cos C \):
\[ \tan A \tan B + \tan B \tan C + \tan C \tan A = \frac{3 + 1 - 2q}{q} = \frac{4 - 2q}{q} \]

5. **Forming the Cubic Equation**:
The general form of a cubic equation with roots \( r_1, r_2, r_3 \) is:
\[ x^3 - (r_1 + r_2 + r_3)x^2 + (r_1r_2 + r_2r_3 + r_3r_1)x - r_1r_2r_3 = 0 \]
Substituting the values we found:
\[ x^3 - \left(\frac{p}{q}\right)x^2 + \left(\frac{4 - 2q}{q}\right)x - \left(\frac{p}{q}\right) = 0 \]

6. **Final Form**:
Multiplying through by \( q \) to eliminate the denominators:
\[ qx^3 - px^2 + (4 - 2q)x - p = 0 \]

### Final Answer:
The cubic equation having roots \( \tan A, \tan B, \tan C \) is:
\[ qx^3 - px^2 + (4 - 2q)x - p = 0 \]