What is the value of $$\frac{[1 - \tan (90 - \theta) + \sec (90 - \theta)]}{[\tan (90 - \theta) + \sec (90 - \theta) + 1]}$$?

$$\frac{[1-\tan(90-\theta)+\sec(90-\theta)]}{[\tan(90-\theta)+\sec(90-\theta)+1]}$$

$$=\frac{[1-\cot\theta\ +\operatorname{cosec}\theta]}{[\cot\theta+\operatorname{cosec}\theta+1]}\ .$$

$$=\frac{\sin\theta\ -\cos\theta\ +1}{\sin\theta\ +1+\cos\theta\ }\ .$$

Now, 

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

And, 

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

So, 

$$\frac{[1-\tan(90-\theta)+\sec(90-\theta)]}{[\tan(90-\theta)+\sec(90-\theta)+1]}$$

$$=\frac{2\tan\frac{\theta}{2}\left(1+\tan\frac{\theta}{2}\right)}{2\left(\tan\frac{\theta}{2}+1\right)}\ .$$

$$=\tan\frac{\theta}{2}\ .$$

B is correct choice.