IF `(sec^2A)hati+hatj+hatk, hati+(sec^2 B)hatj+hatk and hati+hatj+(sec^2 C)hatk` are coplanar then `cot^2 A+cot^2B+cot^2C` is

Options

A. equal to 1
B. equal to 2
C. equal to 0
D. not defined

Answer

To determine the value of \( \cot^2 A + \cot^2 B + \cot^2 C \) given that the vectors \( \vec{a} = (\sec^2 A) \hat{i} + \hat{j} + \hat{k} \), \( \vec{b} = \hat{i} + (\sec^2 B) \hat{j} + \hat{k} \), and \( \vec{c} = \hat{i} + \hat{j} + (\sec^2 C) \hat{k} \) are coplanar, we will follow these steps:

### Step 1: Set up the determinant for coplanarity
The vectors are coplanar if the determinant of the matrix formed by their components is zero. Thus, we need to compute the determinant:

\[ \begin{vmatrix} \sec^2 A & 1 & 1 \\ 1 & \sec^2 B & 1 \\ 1 & 1 & \sec^2 C \end{vmatrix} = 0 \]

### Step 2: Expand the determinant
Using the determinant expansion, we can write:

\[ \sec^2 A \begin{vmatrix} \sec^2 B & 1 \\ 1 & \sec^2 C \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 1 & \sec^2 C \end{vmatrix} + 1 \begin{vmatrix} 1 & \sec^2 B \\ 1 & 1 \end{vmatrix} = 0 \]

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} \sec^2 B & 1 \\ 1 & \sec^2 C \end{vmatrix} = \sec^2 B \cdot \sec^2 C - 1 \)
2. \( \begin{vmatrix} 1 & 1 \\ 1 & \sec^2 C \end{vmatrix} = \sec^2 C - 1 \)
3. \( \begin{vmatrix} 1 & \sec^2 B \\ 1 & 1 \end{vmatrix} = 1 - \sec^2 B \)

Substituting these back into the determinant gives:

\[ \sec^2 A (\sec^2 B \sec^2 C - 1) - (\sec^2 C - 1) + (1 - \sec^2 B) = 0 \]

### Step 3: Simplify the equation
Rearranging the equation yields:

\[ \sec^2 A \sec^2 B \sec^2 C - \sec^2 A - \sec^2 C + 1 + 1 - \sec^2 B = 0 \]

This simplifies to:

\[ \sec^2 A \sec^2 B \sec^2 C - \sec^2 A - \sec^2 B - \sec^2 C + 2 = 0 \]

### Step 4: Use the identity for secant
Recall that \( \sec^2 \theta = 1 + \tan^2 \theta \). Thus, we can substitute:

\[ (1 + \tan^2 A)(1 + \tan^2 B)(1 + \tan^2 C) - (1 + \tan^2 A) - (1 + \tan^2 B) - (1 + \tan^2 C) + 2 = 0 \]

### Step 5: Expand and collect terms
Expanding the left-hand side gives:

\[ 1 + \tan^2 A + \tan^2 B + \tan^2 C + \tan^2 A \tan^2 B + \tan^2 A \tan^2 C + \tan^2 B \tan^2 C + \tan^2 A \tan^2 B \tan^2 C - 3 - (\tan^2 A + \tan^2 B + \tan^2 C) + 2 = 0 \]

This simplifies to:

\[ \tan^2 A \tan^2 B \tan^2 C + \tan^2 A \tan^2 B + \tan^2 A \tan^2 C + \tan^2 B \tan^2 C = 0 \]

### Step 6: Divide by \( \tan^2 A \tan^2 B \tan^2 C \)
Dividing through by \( \tan^2 A \tan^2 B \tan^2 C \) gives:

\[ 1 + \frac{1}{\tan^2 A} + \frac{1}{\tan^2 B} + \frac{1}{\tan^2 C} = 0 \]

### Step 7: Relate to cotangent
This implies:

\[ \cot^2 A + \cot^2 B + \cot^2 C + 1 = 0 \]

Thus:

\[ \cot^2 A + \cot^2 B + \cot^2 C = -1 \]

### Conclusion
Since the sum of squares cannot be negative, we conclude that \( \cot^2 A + \cot^2 B + \cot^2 C \) is not defined.