Question:

Using the identity $\cos(A-B) = \cos A \cos B + \sin A \sin B$, we can write: <br> $D = \begin{vmatrix} 1 & \cos\beta\cos\alpha + \sin\beta\sin\alpha & \cos\gamma\cos\alpha + \sin\gamma\sin\alpha \\ \cos\alpha\cos\beta + \sin\alpha\sin\beta & 1 & \cos\gamma\cos\beta + \sin\gamma\sin\beta \\ \cos\alpha\cos\gamma + \sin\alpha\sin\gamma & \cos\beta\cos\gamma + \sin\beta\sin\gamma & 1 \end{vmatrix}$ <br> Replacing $1$ with $\cos^2 \theta + \sin^2 \theta$: <br> $D = \begin{vmatrix} \cos^2\alpha + \sin^2\alpha & \cos\beta\cos\alpha + \sin\beta\sin\alpha & \cos\gamma\cos\alpha + \sin\gamma\sin\alpha \\ \cos\alpha\cos\beta + \sin\alpha\sin\beta & \cos^2\beta + \sin^2\beta & \cos\gamma\cos\beta + \sin\gamma\sin\beta \\ \cos\alpha\cos\gamma + \sin\alpha\sin\gamma & \cos\beta\cos\gamma + \sin\beta\sin\gamma & \cos^2\gamma + \sin^2\gamma \end{vmatrix}$ <br> This can be written as a product of two determinants: <br> $D = \begin{vmatrix} \cos\alpha & \sin\alpha & 0 \\ \cos\beta & \sin\beta & 0 \\ \cos\gamma & \sin\gamma & 0 \end{vmatrix} \times \begin{vmatrix} \cos\alpha & \cos\beta & \cos\gamma \\ \sin\alpha & \sin\beta & \sin\gamma \\ 0 & 0 & 0 \end{vmatrix}$ <br> Since both determinants have a column or row of zeros, their values are 0. <br> Thus, $D = 0 \times 0 = 0$.