The value of $$\frac{(1+\sin A)(\tan A - \sec A + 1)\sec A}{\tan A + \sec A - (\sec^2 A - \tan^2 A)} \text{ is:}$$​ 

Correct option is B

Given:

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

Formula Used:

​$$\sec^2 A-\tan^2 A=1$$​​

​$$\sin^2 A=1-\cos^2 A$$​​

Solution:
Since, $$\sec A\pm\tan A=\dfrac{1\pm\sin A}{\cos A}$$

now,

​$$=\frac{(1+\sin A)\left(1-(\sec A-\tan A)\right)\frac{1}{\cos A}}{\frac{1+\sin A}{\cos A}-1} \\ \ \\=\frac{(1+\sin A)\left(1-\frac{1-\sin A}{\cos A}\right)\frac{1}{\cos A}}{\frac{1+\sin A-\cos A}{\cos A}}$$​​

​$$=\frac{(1+\sin A)(\sin A+\cos A-1)}{\cos^2 A}\div\frac{1+\sin A-\cos A}{\cos A}$$​​

​$$=\frac{(1+\sin A)(\sin A+\cos A-1)}{\cos A (1+\sin A-\cos A)}.$$​

​$$= \frac{\sin^2 A+\cos A+\sin A\cos A-1}{\cos A(1+\sin A-\cos A)}$$​​

= $$\frac{1-\cos^2 A+\cos A+\sin A\cos A-1}{\cos A(1+\sin A-\cos A)}$$​

Therefore, the expression simplifies to

​$$\frac{\cos A(1+\sin A-\cos A)}{\cos A(1+\sin A-\cos A)}=1.$$​​