[กิตติ]

จาก $(\sec^2A-\tan^2A)^2=1$
$1=\sec^2A\tan^2A$
$\tan^4A+\tan^2A-1=0$
$\tan^2A=\frac{\sqrt{5}-1 }{2} $
$\cot^2A=\frac{\sqrt{5}+1 }{2}$

จาก $(\csc^2A-\cot^2A)^2+2 \csc^2A \cot^2A=\csc^4A+\cot^4A$
$\csc^4A+\cot^4A=1+2 \csc^2A \cot^2A$
$=1+2(1+\cot^2A)\cot^2A$
$=1+2\cot^2A+2\cot^4A$
$=1+(\sqrt{5}+1)+(3+\sqrt{5})$
$=5+2\sqrt{5}$