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 solve the problem, we need to determine the value of \( \cot^2 A + \cot^2 B + \cot^2 C \) given that the vectors

\[ \vec{v_1} = (\sec^2 A) \hat{i} + \hat{j} + \hat{k}, \]
\[ \vec{v_2} = \hat{i} + (\sec^2 B) \hat{j} + \hat{k}, \]
\[ \vec{v_3} = \hat{i} + \hat{j} + (\sec^2 C) \hat{k} \]

are coplanar.

### Step 1: Set up the determinant for coplanarity
For three vectors to be coplanar, the scalar triple product (or the determinant of the matrix formed by these vectors) must be 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
We can expand the determinant using the formula for a 3x3 matrix:

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

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:

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

### Step 3: Set the determinant equal to zero
Setting the determinant equal to zero:

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

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: Substitute \( \sec^2 \) with \( 1 + \tan^2 \)
Using the identity \( \sec^2 \theta = 1 + \tan^2 \theta \):

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

Expanding this gives:

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

### Step 5: Simplify the equation
After simplification, we get:

\[ \tan^2 A \tan^2 B + \tan^2 B \tan^2 C + \tan^2 C \tan^2 A = 1 \]

### Step 6: Use the cotangent identity
Using the identity \( \cot^2 \theta = \frac{1}{\tan^2 \theta} \):

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

This implies:

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

### Final Answer
Thus, the value of \( \cot^2 A + \cot^2 B + \cot^2 C \) is:

\[ \boxed{1} \]