To prove that prove that $cos^8 \theta sec^6 \alpha , \frac{1}{2 } ,sin^8 \theta cosec^6 \alpha $ are in A.P If $\cos^4 \theta \sec^2 \alpha , \frac{1}{2 } ,\sin^4 \theta \csc^2 \alpha $ are in A.P , 
then prove that $\cos^8 \theta \sec^6 \alpha , \frac{1}{2 } ,\sin^8 \theta \csc^6  \alpha $ are in A.P
Now i have reached upto 
$1=\cos^4 \theta \sec^2 \alpha + \sin^4 \theta \csc^2 \alpha$.
By completing square i have $(\sin^2 \theta \cot\alpha - \cos^2 \theta \tan \alpha)^2=0$ so i get $\tan \theta= \pm \tan \alpha$
 How do i proceed?
Thanks

You have done rightly. Now we can proceed as follows: $$\tan \theta = \pm \tan \alpha \Rightarrow \tan^2 \theta = \tan^2 \alpha  \Rightarrow \sec^2 \theta -1 =\sec^2 \alpha -1 \Rightarrow \sec^2 \theta = \sec^2 \alpha \Leftrightarrow \sec^6 \theta = \sec^6 \alpha \tag{1}$$
Also, $$\tan^2 \theta = \tan^2 \alpha \Rightarrow \cot^2 \theta = \cot^2 \alpha \Rightarrow \csc^2 \theta -1 = \csc^2 \alpha -1 \Rightarrow \csc^2 \theta = \csc^2 \alpha \Leftrightarrow \csc^6 \theta = \csc^6 \alpha \tag {2}$$
Hope you can take it from here by using $(1)$ and $(2)$. 
