Proving that two determinants are equal without expanding them So I need to prove that 

$$
    \begin{vmatrix}
    \sin^2(\alpha) & \cos(2\alpha) & \cos^2(\alpha) \\
    \sin^2(\beta) & \cos(2\beta) & \cos^2(\beta) \\
    \sin^2(\gamma) & \cos(2\gamma) & \cos^2(\gamma) \\
    \end{vmatrix}
$$
$$
= \begin{vmatrix}
    \sin(\alpha) & \cos(\alpha) & \sin(\alpha + \delta) \\
    \sin(\beta) & \cos(\beta) & \sin(\beta + \delta) \\
    \sin(\gamma) & \cos(\gamma) & \sin(\gamma + \delta) \\
    \end{vmatrix}
$$

Now, 
$$
  \begin{vmatrix}
    \sin^2(\alpha) & \cos(2\alpha) & \cos^2(\alpha) \\
    \sin^2(\beta) & \cos(2\beta) & \cos^2(\beta) \\
    \sin^2(\gamma) & \cos(2\gamma) & \cos^2(\gamma) \\
    \end{vmatrix} = \begin{vmatrix}
    \sin^2(\alpha) & \cos^2(\alpha) - \sin^2(\alpha) & \cos^2(\alpha) \\
    \sin^2(\beta) & \cos^2(\beta) - \sin^2(\beta) & \cos^2(\beta) \\
    \sin^2(\gamma) & \cos^2(\gamma) - \sin^2(\gamma) & \cos^2(\gamma) \\
    \end{vmatrix}
$$
Adding column $1$ to column $2$ then makes column $2$ and column $3$ equal and hence the first determinant $ = 0$.
I'm stuck in trying to prove that the second determinant is also zero, so I need help in that.

HINT: Relate the third column to the first two using the trigonometric identity
$$\sin(\theta+\delta)=\sin\theta\cos\delta+\cos\theta\sin\delta.$$
