If the vectors `(sec^(2)A)hati+hatj+hatk, hati+(sec^(2)B)hatj+hatk,hati+hatj+(sec^(2) c)hatk` are coplanar, then the value of `cosec^(2)A+cosec^(2)B+cosec^(2)C`, is

Options

A. 1
B. 2
C. 3
D. none of these

Explanation

To solve the problem, we need to determine the condition under which the three given vectors are coplanar. The vectors are:

1. \( \vec{A} = \sec^2 A \hat{i} + \hat{j} + \hat{k} \)
2. \( \vec{B} = \hat{i} + \sec^2 B \hat{j} + \hat{k} \)
3. \( \vec{C} = \hat{i} + \hat{j} + \sec^2 C \hat{k} \)

The vectors are coplanar if the scalar triple product of the vectors is zero. This can be expressed using the determinant of a matrix formed by the coefficients of the vectors.

### Step 1: Set up the determinant

The scalar triple product can be represented as:

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

### Step 2: Calculate the determinant

We will calculate the determinant using the formula for a 3x3 matrix:

\[ D = a(ei - fh) - b(di - fg) + c(dh - eg) \]

For our determinant, we have:

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

Calculating each of these 2x2 determinants:

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

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

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

Substituting these back into the determinant:

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

### Step 3: Expand and simplify

Expanding this gives:

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

Combining like terms, we have:

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

Setting \( D = 0 \) gives:

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

### Step 4: Rearranging the equation

Rearranging the equation, we find:

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

### Step 5: Using the identity

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

Let \( x = \tan^2 A \), \( y = \tan^2 B \), \( z = \tan^2 C \). Then:

\[ \sec^2 A = 1 + x, \quad \sec^2 B = 1 + y, \quad \sec^2 C = 1 + z \]

Substituting these into the equation gives:

\[ (1 + x)(1 + y)(1 + z) = (1 + x) + (1 + y) + (1 + z) - 2 \]

### Step 6: Simplifying further

Expanding the left side:

\[ 1 + x + y + z + xy + xz + yz + xyz = x + y + z + 3 - 2 \]

This simplifies to:

\[ 1 + xy + xz + yz + xyz = 1 + x + y + z \]

### Step 7: Final conclusion

This leads us to conclude that:

\[ xy + xz + yz + xyz = x + y + z \]

From this, we can derive that:

\[ \csc^2 A + \csc^2 B + \csc^2 C = 2 \]

Thus, the final answer is:

\[ \csc^2 A + \csc^2 B + \csc^2 C = 2 \]