If $$\frac{(1 + \sin \theta - \cos \theta)}{(1 + \sin \theta + \cos \theta)} + \frac{(1 + \sin \theta + \cos \theta)}{(1 + \sin \theta - \cos \theta)} = 4$$, then which of the following values will be suitable for $$\theta$$

$$\frac{(1+\sin\theta-\cos\theta)}{(1+\sin\theta+\cos\theta)}+\frac{(1+\sin\theta+\cos\theta)}{(1+\sin\theta-\cos\theta)}=4$$

$$=$$>  $$\frac{(1+\sin\theta-\cos\theta)^2+(1+\sin\theta+\cos\theta)^2}{(1+\sin\theta+\cos\theta)(1+\sin\theta-\cos\theta)}=4$$

$$=$$>  $$\frac{(1+\sin\theta)^2+\cos^2\theta\ -2\left(1+\sin\theta\ \right)\cos\theta\ +(1+\sin\theta)^2+\cos^2\theta\ +2\left(1+\sin\theta\right)\cos\theta\ \ }{\left(1+\sin\theta\right)^2-\cos^2\theta}=4$$

$$=$$>  $$\frac{2(1+\sin^2\theta+2\sin\theta\ )+2\cos^2\theta}{1+\sin^2\theta\ +2\sin\theta\ -\cos^2\theta\ }\ =4$$

$$=$$>  $$2+2\sin^2\theta+4\sin\theta\ +2\cos^2\theta=4+4\sin^2\theta\ +8\sin\theta\ -4\cos^2\theta\ $$

$$=$$>  $$2+2\left(\sin^2\theta+\cos^2\theta\ \right)=4+4\sin^2\theta\ +4\sin\theta\ -4\left(1-\sin^2\theta\ \right)$$

$$=$$>  $$2+2=4+4\sin^2\theta\ +4\sin\theta\ -4\left(1-\sin^2\theta\ \right)$$

$$=$$>  $$0=4\sin^2\theta\ +4\sin\theta\ -4+4\sin^2\theta\ $$

$$=$$>  $$8\sin^2\theta\ +4\sin\theta\ -4=0$$

$$=$$>  $$2\sin^2\theta\ +\sin\theta\ -1=0$$

$$=$$>  $$2\sin^2\theta\ +2\sin\theta\ -\sin\theta\ -1=0$$

$$=$$>  $$2\sin\theta\left(\sin\theta\ +1\right)\ -1\left(\sin\theta\ +1\right)=0$$

$$=$$>  $$\left(\sin\theta\ +1\right)\left(2\sin\theta\ -1\right)=0$$

$$=$$>  $$\sin\theta\ +1=0$$   or   $$2\sin\theta\ -1=0$$

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

$$\sin\theta$$ cannot be negative

$$=$$>  $$\sin\theta\ =\frac{1}{2}$$

$$=$$>  $$\theta\ =30^{\circ\ }$$

Hence, the correct answer is Option D