<s>[INST] 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
 [/INST] We have $\cos^4\theta\sec^2\alpha+\sin^4\theta\csc^2\alpha=1$
Set $\cos^2\theta=a,\sin^4\theta=(1-a)^2$ to solve for $a$ to find $a=\cos^2\alpha$
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