Determinants

By using properties of determinants prove that the determinant $\begin{vmatrix} a & \sin x & \cos x \\ -\sin x & -a & 1 \\ \cos x & 1 & a \end{vmatrix}$ is independent of x.

Let $\Delta =$ $\begin{vmatrix} a & \sin x & \cos x \\ -\sin x & -a & 1 \\ \cos x & 1 & a \end{vmatrix}$

$C_1 \rightarrow C_1 + C_2 \implies \Delta =$ $\begin{vmatrix} a+\sin x & \sin x & \cos x \\ -a-\sin x & -a & 1 \\ 1+\cos x & 1 & a \end{vmatrix}$

$R_1 \rightarrow R_1 + R_2, R_2 \rightarrow R_2+aR_3 \implies$

$\Delta =$ $\begin{vmatrix} 0 & -a+\sin x & 1+\cos x \\ a\cos x-\sin x & 0 & 1+a^2 \\ 1+\cos x & 1 & a \end{vmatrix}$

Expanding along $R_1$ gives $\begin{aligned} \Delta = & -\left(-a+\sin x\right) \left[a\left(a\cos x -\sin x\right)-\left(1+a^2\right)\left(1+\cos x\right)\right] + \\ & \left(1+\cos x\right)\left(a\cos x - \sin x\right) \\ \text{i.e. } \Delta = & -\left(-a+\sin x\right)\left(a^2\cos x -a\sin x -1-\cos x -a^2 -a^2 \cos x\right) + \\ & a\cos x - \sin x + a \cos^2 x - \sin x \cos x \\ \text{i.e. } \Delta = & \left(-a+\sin x\right) \left(a\sin x + \cos x +1 +a^2\right) \\ & a\cos x - \sin x + a \cos^2 x - \sin x \cos x \\ \text{i.e. } \Delta = & -a^2 \sin x - a\cos x -a - a^3 +a \sin^2 x +\sin x \cos x +\sin x + \\ & a^2 \sin x + a\cos x - \sin x + a \cos^2 x - \sin x \cos x \\ \text{i.e. } \Delta = & -a^3 \text{ which is independent of x.} \end{aligned}$