Statement-1: The system of linear equations
$$x + (\sin\alpha)y + (\cos\alpha)z = 0$$
$$x + (\cos\alpha)y + (\sin\alpha)z = 0$$
$$x - (\sin\alpha)y - (\cos\alpha)z = 0$$
has a non-trivial solution for only one value of $$\alpha$$ lying in the interval $$(0, \frac{\pi}{2})$$.
Statement-2: The equation in $$\alpha$$ $$\begin{vmatrix} \cos\alpha & \sin\alpha & \cos\alpha \\ \sin\alpha & \cos\alpha & \sin\alpha \\ \cos\alpha & -\sin\alpha & \cos\alpha \end{vmatrix} = 0$$ has only one solution lying in the interval $$(0, \frac{\pi}{2})$$.

Statement 1:

$$D = \begin{vmatrix} 1 & \sin \alpha & \cos \alpha \\ 1 & \cos \alpha & \sin \alpha \\ 1 & -\sin \alpha & -\cos \alpha \end{vmatrix}$$

Using row operations ($$R_1 \to R_1 + R_3$$):

$$D = \begin{vmatrix} 2 & 0 & 0 \\ 1 & \cos \alpha & \sin \alpha \\ 1 & -\sin \alpha & -\cos \alpha \end{vmatrix}$$

$$D = 2(-\cos^2 \alpha + \sin^2 \alpha) = -2(\cos^2 \alpha - \sin^2 \alpha) = -2\cos(2\alpha)$$

$$-2\cos(2\alpha) = 0 \implies \cos(2\alpha) = 0$$

$$2\alpha = \frac{\pi}{2} \implies \alpha = \frac{\pi}{4}$$

In the interval $$(0, \frac{\pi}{2})$$, there is only one value ($$\alpha = \frac{\pi}{4}$$). Statement-1 is True.

Statement 2:

$$\begin{vmatrix} \cos \alpha & \sin \alpha & \cos \alpha \\ \sin \alpha & \cos \alpha & \sin \alpha \\ \cos \alpha & -\sin \alpha & \cos \alpha \end{vmatrix} = 0$$

Since the determinant is $$0 = 0$$ (two identical columns), the equation is an identity. It has infinitely many solutions in the interval $$(0, \frac{\pi}{2})$$, not just one. Statement-2 is False.