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

$=\sec ^{2} \theta \cdot \cos ^{2} \theta$ $\left[\because 1+\tan ^{2} \theta=\sec ^{2} \theta\right.$ and $\left.\cos ^{2} \theta+\sin ^{2} \theta=1\right]$

$=\frac{1}{\cos ^{2} \theta} \cdot \cos ^{2} \theta=1$ $\left[\because \sec \theta=\frac{1}{\cos \theta}\right]$