The expression $$\frac{\tan A}{1 - \cot A} + \frac{\cot A}{1 - \tan A}$$ can be written as :

We have to simplify the expression

$$\frac{\tan A}{1-\cot A}+\frac{\cot A}{1-\tan A}$$

To remove the mixed trigonometric functions, we first recall the basic identities

$$\tan A=\frac{\sin A}{\cos A},\qquad \cot A=\frac{\cos A}{\sin A}.$$

Substituting these in the given expression, we treat each fraction separately.

For the first fraction:

$$\frac{\tan A}{1-\cot A} =\frac{\dfrac{\sin A}{\cos A}} {1-\dfrac{\cos A}{\sin A}}.$$

Combining the terms in the denominator we get

$$1-\frac{\cos A}{\sin A} =\frac{\sin A-\cos A}{\sin A}.$$

So the first fraction becomes

$$\frac{\dfrac{\sin A}{\cos A}} {\dfrac{\sin A-\cos A}{\sin A}} =\frac{\sin A}{\cos A}\times\frac{\sin A}{\sin A-\cos A} =\frac{\sin^2 A}{\cos A(\sin A-\cos A)}.$$

Now, for the second fraction:

$$\frac{\cot A}{1-\tan A} =\frac{\dfrac{\cos A}{\sin A}} {1-\dfrac{\sin A}{\cos A}}.$$

The denominator simplifies as

$$1-\frac{\sin A}{\cos A} =\frac{\cos A-\sin A}{\cos A}.$$

Thus the second fraction is

$$\frac{\dfrac{\cos A}{\sin A}} {\dfrac{\cos A-\sin A}{\cos A}} =\frac{\cos A}{\sin A}\times\frac{\cos A}{\cos A-\sin A} =\frac{\cos^2 A}{\sin A(\cos A-\sin A)}.$$

The entire expression is therefore

$$\frac{\sin^2 A}{\cos A(\sin A-\cos A)} +\frac{\cos^2 A}{\sin A(\cos A-\sin A)}.$$

Notice that $$\cos A-\sin A=-(\sin A-\cos A).$$ We can rewrite the second term by inserting this minus sign:

$$\frac{\cos^2 A}{\sin A(\cos A-\sin A)} =-\frac{\cos^2 A}{\sin A(\sin A-\cos A)}.$$

Hence the sum becomes

$$\frac{\sin^2 A}{\cos A(\sin A-\cos A)} -\frac{\cos^2 A}{\sin A(\sin A-\cos A)}.$$

We now take the common denominator $$\sin A\cos A(\sin A-\cos A)$$ and combine the numerators:

$$ \frac{\sin^2 A}{\cos A(\sin A-\cos A)} =\frac{\sin^3 A}{\sin A\cos A(\sin A-\cos A)},$$

$$ -\frac{\cos^2 A}{\sin A(\sin A-\cos A)} =-\frac{\cos^3 A}{\sin A\cos A(\sin A-\cos A)}.$$

Adding these gives

$$\frac{\sin^3 A-\cos^3 A} {\sin A\cos A(\sin A-\cos A)}.$$

Next, we use the algebraic identity for the difference of cubes:

$$a^3-b^3=(a-b)(a^2+ab+b^2).$$

Putting $$a=\sin A,\; b=\cos A$$ we obtain

$$\sin^3 A-\cos^3 A =(\sin A-\cos A)(\sin^2 A+\sin A\cos A+\cos^2 A).$$

But $$\sin^2 A+\cos^2 A=1,$$ so

$$\sin^2 A+\sin A\cos A+\cos^2 A =1+\sin A\cos A.$$

Thus the numerator simplifies to

$$(\sin A-\cos A)(1+\sin A\cos A).$$

Our fraction is now

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

The factor $$\sin A-\cos A$$ appears in both numerator and denominator, so it cancels out (assuming $$\sin A\neq\cos A$$):

$$\frac{1+\sin A\cos A}{\sin A\cos A}.$$

We separate the numerator:

$$\frac{1}{\sin A\cos A}+\frac{\sin A\cos A}{\sin A\cos A} =\frac{1}{\sin A\cos A}+1.$$

The reciprocal identities $$\sec A=\dfrac{1}{\cos A}$$ and $$\csc A=\dfrac{1}{\sin A}$$ give

$$\frac{1}{\sin A\cos A} =\frac{1}{\sin A}\times\frac{1}{\cos A} =\csc A\sec A.$$

Therefore the entire expression equals

$$\csc A\sec A+1.$$

This matches option D.

Hence, the correct answer is Option D.