Correct Answer: 1

Solution : $\frac{\sin \theta+\cos \theta-1}{\sin \theta-\cos \theta+1} \times \sqrt{\frac{1+\sin \theta}{1-\sin \theta}}$
= $\frac{\sin \theta+\cos \theta-1}{\sin \theta-\cos \theta+1}\times\frac{\sin \theta-\cos \theta-1}{\sin \theta-\cos \theta-1} \times \sqrt{\frac{1+\sin \theta}{1-\sin \theta}\times\frac{1+\sin \theta}{1+\sin \theta}}$
= $\frac{(\sin \theta-1)^2-\cos^2 \theta}{(\sin \theta-\cos \theta)^2-1^2} \times \sqrt{\frac{(1+\sin \theta)^2}{1^2-\sin^2 \theta}}$
= $\frac{\sin^2 \theta+1-2\sin\theta-\cos^2 \theta}{\sin^2 \theta+\cos^2 \theta-2\sin\theta\cos\theta-1} \times \sqrt{\frac{(1+\sin \theta)^2}{\cos^2 \theta}}$
We know that,
$\sin^2\theta+\cos^2\theta=1$
= $\frac{1-\cos^2 \theta+1-2\sin\theta-\cos^2 \theta}{1-2\sin\theta\cos\theta-1} \times \frac{1+\sin \theta}{\cos\theta}$
= $\frac{2-2\cos^2 \theta-2\sin\theta}{-2\sin\theta\cos\theta} \times \frac{1+\sin \theta}{\cos\theta}$
= $\frac{2(1-\cos^2 \theta-\sin\theta)}{-2\sin\theta\cos\theta} \times \frac{1+\sin \theta}{\cos\theta}$
= $\frac{\sin^2 \theta-\sin\theta}{-\sin\theta\cos\theta} \times \frac{1+\sin \theta}{\cos\theta}$
= $\frac{\sin \theta(\sin\theta-1)}{-\sin\theta\cos\theta} \times \frac{1+\sin \theta}{\cos\theta}$
= $\frac{(\sin\theta-1)}{-\cos\theta} \times \frac{1+\sin \theta}{\cos\theta}$
= $(\frac{\sin \theta}{-\cos\theta}+\frac{1}{\cos\theta}) \times (\frac{1}{\cos\theta}+\frac{\sin}{\cos\theta})$
= $(-\tan\theta+\sec\theta)\times(\sec\theta+\tan\theta)$
= $(-\tan^2\theta+\sec^2\theta)$
= 1
Hence, the correct answer is 1.