2010 AMC 12B Problem 20

The common ratio is $r=\dfrac{a_2}{a_1}=\cot x.$ Then $a_4=a_3\cdot r=\tan x\cot x=1.$ From $a_3=a_1r^2,$ we get $\tan x=\sin x\cot^2 x=\dfrac{\cos^2 x}{\sin x},$ so $\sin^2 x=\cos^3 x,$ i.e. $(\cos^2 x)(1+\cos x)=1.$ Hence $1+\cos x=\dfrac{1}{\cos^2 x}.$ Also $r^2=\dfrac{\cos^2 x}{\sin^2 x}=\dfrac{\cos^2 x}{\cos^3 x}=\dfrac{1}{\cos x},$ so $r^4=\dfrac{1}{\cos^2 x}=1+\cos x.$ Therefore $1+\cos x=a_4\cdot r^4=a_8,$ so $n=8.$ Thus, the correct answer is E.