the value of the determinant $\left|\begin{array}{l}\sin A\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\cos A\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\sin (A+\theta )\\ \sin B\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\cos B\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\sin (B+\theta )\\ \sin C\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\cos C\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\sin (C+\theta )\end{array}\right|$ is independent of

write $\sin (A+\theta )=\sin A\cos \theta +\cos A\sin \theta$ etc.

Now apply $C_3-\cos \theta C_1-\sin \theta C_2\text{ then }{C}_3$ becomes column of zeros. Hence $∆ = 0$ as such it is independent of all A, B, C and $\theta$.