$\cos^8x.\sec^6y,\frac12,\sin^8x.\csc^6y$ in AP if $\cos^4x.\sec^2y,\frac12,\sin^4x.\csc^2y$ in A.P

Question

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 ?

Answer 1

Hint:

$$\dfrac12-\cos^4x=2\left(\sin^4x-\dfrac12\right)$$

$$\iff2-(1+\cos2x)^2=2(1-\cos2x)^2-4$$

Solve for $\cos2x$

$$\csc^2y-\dfrac12=2\left(\dfrac12-\sec^2y\right)$$

$$\iff2-4(1+\tan^2y)=2(1+\cot^2y)-1$$

$$4\tan^2y+2\cdot\dfrac1{\tan^2y}-5=0$$

Solve for $\tan^2y\ge0$

Answer 2

Substitute $\sin^2y=1-\cos^2y$ and $\sin^2x=1-\cos^2x$, and factor the result.