The question should be:
prove (cos(3x)-cos(7x))/(sin(3x)+sin(7x))=tan(2x).
cos(A+B)=cosAcosB-sinAsinB
cos(A-B)=cosAcosB+sinAsinB
So, cos(A-B)-cos(A+B)=2sinAsinB and
cos(A+B)+cos(A-B)=2cosAcosB.
Also, sin(A+B)=sinAcosB+cosAsinB and
sin(A-B)=sinAcosB-cosAsinB, so
sin(A+B)+sin(A-B)=2sinAcosB
sin(A+B)-sin(A-B)=2cosAcosB
If A-B=3x and A+B=7x, 2A=10x, A=5x and B=2x.
Therefore, cos(3x)-cos(7x)=2sin(5x)sin(2x) and
sin(3x)+cos(7x)=2sin5xcos(2x) and
(cos(3x)-cos(7x))/(sin(3x)+sin(7x)=
2sin(5x)sin(2x)/(2sin(5x)cos(2x))=tan(2x) QED