If $\cos^4x.\sec^2y,\dfrac{1}{2},\sin^4x.\csc^2y$ are in A.P, then prove that $\cos^8x.\sec^6y,\dfrac{1}{2},\sin^8x.\csc^6y$ in AP.
My Attempt $$ \cos^4x.\sec^2y+\sin^4.x\csc^2y=\frac{\cos^4x}{\cos^2y}+\frac{\sin^4x}{\sin^2y}=1\\ \implies\sin^2y.\cos^4x+\cos^2y.\sin^4x=\sin^2y.\cos^2y $$ $$ \cos^8x.\sec^6y+\sin^8x.\csc^6y=\frac{\cos^8x}{\cos^6y}+\frac{\sin^8x}{\sin^6y} $$ How do I know that the given terms are in A.P, G.P or H.P ?