L.H.S
sinA+sin5A+sin9AcosA+cos5A+cos9A
We know that
sinC+sinD=2sin(C+D2)⋅cos(C−D2)
cosC+cosD=2cos(C+D2)⋅cos(C−D2)
Therefore,
=2sin(A+9A2)⋅cos(A−9A2)+sin5A2cos(A+9A2)⋅cos(A−9A2)+cos5A
=2sin(5A)⋅cos(4A)+sin5A2cos(5A)⋅cos(4A)+cos5A
=2cos(4A)+12cos(4A)+1(sin5Acos5A)
=sin5Acos5A
=tan5A
Hence, proved.