Question:
$\sec ^{2} \theta-\frac{\sin ^{2} \theta-2 \sin ^{4} \theta}{2 \cos ^{4} \theta-\cos ^{2} \theta}=1$
Solution:
$\sec ^{2} \theta-\frac{\sin ^{2} \theta-2 \sin ^{4} \theta}{2 \cos ^{4} \theta-\cos ^{2} \theta}$
$=\sec ^{2} \theta-\frac{\sin ^{2} \theta\left(1-2 \sin ^{2} \theta\right)}{\cos ^{2} \theta\left(2 \cos ^{2} \theta-1\right)}$
$=\sec ^{2} \theta-\frac{\sin ^{2} \theta\left(1-2 \sin ^{2} \theta\right)}{\cos ^{2} \theta\left[2\left(1-\sin ^{2} \theta\right)-1\right]} \quad\left[\cos ^{2} \theta+\sin ^{2} \theta=1\right]$
$=\sec ^{2} \theta-\frac{\sin ^{2} \theta\left(1-2 \sin ^{2} \theta\right)}{\cos ^{2} \theta\left(1-2 \sin ^{2} \theta\right)}$
$=\sec ^{2} \theta-\tan ^{2} \theta$
$=1 \quad\left(1+\tan ^{2} \theta=\sec ^{2} \theta\right)$