For all values of A,B,C and P,Q,R the determinant given below
`|(cos(A-P),cos(A-Q),cos(A-R)),(cos(B-P),cos(B-Q),cos(B-R)),(cos(C-P),cos(C-Q),cos(C-R))|` is

Options

A. `cos A cos B cos C`
B. `sin P sinQ sinR`
C. 0
D. `sum sin(A+P)`

Answer

To solve the determinant

\[ D = \begin{vmatrix} \cos(A-P) & \cos(A-Q) & \cos(A-R) \\ \cos(B-P) & \cos(B-Q) & \cos(B-R) \\ \cos(C-P) & \cos(C-Q) & \cos(C-R) \end{vmatrix} \]

we will use the cosine subtraction formula, which states:

\[ \cos(x - y) = \cos x \cos y + \sin x \sin y \]

### Step 1: Apply the cosine subtraction formula

We can rewrite each element of the determinant using the cosine subtraction formula:

- For the first row:
- \(\cos(A-P) = \cos A \cos P + \sin A \sin P\)
- \(\cos(A-Q) = \cos A \cos Q + \sin A \sin Q\)
- \(\cos(A-R) = \cos A \cos R + \sin A \sin R\)

- For the second row:
- \(\cos(B-P) = \cos B \cos P + \sin B \sin P\)
- \(\cos(B-Q) = \cos B \cos Q + \sin B \sin Q\)
- \(\cos(B-R) = \cos B \cos R + \sin B \sin R\)

- For the third row:
- \(\cos(C-P) = \cos C \cos P + \sin C \sin P\)
- \(\cos(C-Q) = \cos C \cos Q + \sin C \sin Q\)
- \(\cos(C-R) = \cos C \cos R + \sin C \sin R\)

Thus, we can express the determinant as:

\[ D = \begin{vmatrix} \cos A \cos P + \sin A \sin P & \cos A \cos Q + \sin A \sin Q & \cos A \cos R + \sin A \sin R \\ \cos B \cos P + \sin B \sin P & \cos B \cos Q + \sin B \sin Q & \cos B \cos R + \sin B \sin R \\ \cos C \cos P + \sin C \sin P & \cos C \cos Q + \sin C \sin Q & \cos C \cos R + \sin C \sin R \end{vmatrix} \]

### Step 2: Factor the determinant

We can factor out the common terms from each column. The determinant can be expressed as a product of two determinants:

\[ D = \begin{vmatrix} \cos A & \cos A & \cos A \\ \cos B & \cos B & \cos B \\ \cos C & \cos C & \cos C \end{vmatrix} \begin{vmatrix} \sin P & \sin P & \sin P \\ \sin Q & \sin Q & \sin Q \\ \sin R & \sin R & \sin R \end{vmatrix} \]

### Step 3: Evaluate the determinants

The first determinant has identical rows, hence its value is 0:

\[ \begin{vmatrix} \cos A & \cos A & \cos A \\ \cos B & \cos B & \cos B \\ \cos C & \cos C & \cos C \end{vmatrix} = 0 \]

Similarly, the second determinant also has identical rows, hence its value is also 0:

\[ \begin{vmatrix} \sin P & \sin P & \sin P \\ \sin Q & \sin Q & \sin Q \\ \sin R & \sin R & \sin R \end{vmatrix} = 0 \]

### Conclusion

Thus, the value of the original determinant \(D\) is:

\[ D = 0 \]