Prove that: $\sin \left(A-B\right)\cos \left(A+B\right)+\sin \left(B-C\right)\cos \left(B+C\right)+\sin \left(C-D\right)\cos \left(C+D\right)+\sin \left(D-A\right)\cos \left(A+D\right)=0$.

Prove that $\sin \left(A+B-2C\right)\cos B-\sin \left(A+C-2B\right)\cos C=\sin \left(B-C\right)\cos \left(B+C-A\right)+\sin \left(B-C\right)\cos \left(C+A-B\right)+\sin \left(B-C\right)\cos \left(A+B-C\right)$.

Prove that $\sin \left(A+B+C+D\right)+\sin \left(A+B-C-D\right)+\sin \left(A+B+C-D\right)+\sin \left(A+B-C+D\right)=4\sin \left(A+B\right)\cos C\cos D$.

Theorem: $\tan A+\tan B+\tan C=\tan A\tan B\tan C$ is true for values of $A, B$, and $C$ if $A+B+C=180°$. Prove that the theorem is still true if we substitute for $A,B \& C$, respectively the quantities.

$\left(1\right)$ $90°-\frac{A}{2}$, $90°-\frac{B}{2}$ and $90°-\frac{C}{2}$

$\left(2\right)$ $180°-2A$, $180°-2B$ and $180°-2C$
 

If $x+y+z=xyz$, then prove that $\frac{3x-x^3}{1-3x^2}+\frac{3y-y^3}{1-3y^2}+\frac{3z-z^3}{1-3z^2}=\frac{3x-x^3}{1-3x^2}· \frac{3y-y^3}{1-3y^2}·\frac{3z-z^3}{1-3z^2}$.

If $x+y+z=xyz$, then prove that $x\left(1-y^2\right)\left(1-z^2\right)+y\left(1-x^2\right)\left(1-z^2\right)+z\left(1-x^2\right)\left(1-y^2\right)=4xyz$.

Solve the equation $\sin \theta +\sin 7\theta =\sin 4\theta$.

Solve the equation $\cos \theta +\cos 7\theta =\cos 4\theta$.