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