Solve this

$(\tan \theta+\sec \theta-1)(\tan \theta+\sec \theta+1)$

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

$=\tan ^{2} \theta+\sec ^{2} \theta+2 \tan \theta \sec \theta-1$

$=2 \tan ^{2} \theta+2 \tan \theta \sec \theta \quad\left(1+\tan ^{2} \theta=\sec ^{2} \theta\right)$

$=2 \tan \theta(\tan \theta+\sec \theta)$

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

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

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

$=\frac{2 \sin \theta(1+\sin \theta)}{(1-\sin \theta)(1+\sin \theta)}$

$=\frac{2 \sin \theta}{1-\sin \theta}$