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

Correct option is C

Given:

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

Solution:

$$\frac{\frac{\sin A}{\cos A}}{1 - \frac{\cos A}{\sin A}} + \frac{\frac{\cos A}{\sin A}}{1 - \frac{\sin A}{\cos A}}$$​​

Denominators;

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

Then,

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

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

Now, 

$$\frac{\sin^3 A - \cos^3 A}{\sin A \cos A (\sin A - \cos A)} \\ \ \\ = \frac{(\sin A - \cos A)(1 + \sin A \cos A) }{\sin A \cos A (\sin A - \cos A)}$$

​$$= \frac{\ 1+ \sin A\cos A}{\sin A \cos A}\\ \ \\ = \sec A \cosec A + 1$$​​

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