Solve this

We know

$\sin ^{2} \theta+\cos ^{2} \theta=1$

Squaring on both sides, we get

$\left(\sin ^{2} \theta+\cos ^{2} \theta\right)^{2}=1$

$\Rightarrow\left(\sin ^{2} \theta\right)^{2}+\left(\cos ^{2} \theta\right)^{2}+2 \sin ^{2} \theta \cos ^{2} \theta=1 \quad\left[(a+b)^{2}=a^{2}+b^{2}+2 a b\right]$

$\Rightarrow \sin ^{4} \theta+\cos ^{4} \theta=1-2 \sin ^{2} \theta \cos ^{2} \theta$

$\therefore \frac{\sin ^{4} \theta+\cos ^{4} \theta}{1-2 \sin ^{2} \theta \cos ^{2} \theta}=\frac{1-2 \sin ^{2} \theta \cos ^{2} \theta}{1-2 \sin ^{2} \theta \cos ^{2} \theta}=1$