​$$\frac{\sin \theta(1+ \cos\theta)}{1+ \cos \theta - \sin^2 \theta)}\times (\sec\theta + \tan \theta)(1- \sin\theta), 0^o < \theta < 90^o$$​ is equal to:

Correct option is D

Given:

​$$\frac{\sin \theta(1+ \cos\theta)}{1+ \cos \theta - \sin^2 \theta)}\times (\sec\theta + \tan \theta)(1- \sin\theta)$$​​

Formula Used:

​$$\tan \theta = \frac{\sin \theta}{\cos\theta}$$​​

​$$\sin^2 \theta + \cos^2 \theta = 1$$​​

​$$\sin( \theta) = \frac{1}{\cosec \theta}$$​​

Solution:​

​$$\frac{\sin \theta(1+ \cos\theta)}{1+ \cos \theta - \sin^2 \theta)}\times (\sec\theta + \tan \theta)(1- \sin\theta)$$​​

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

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

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

= $$\frac{\sin \theta}{\cos \theta}\times \left(\frac{\cos^2 \theta}{\cos \theta}\right)$$​​

​$$=\sin \theta$$​​