The expression `(tanA)/(1-cotA)+(cotA)/(1-tanA)` can be written as

Options

A. `sinA cos A+1`
B. `sec A coses A+1`
C. `tanA+cot A`
D. `sec A+cosec A`

Explanation

To simplify the expression \(\frac{\tan A}{1 - \cot A} + \frac{\cot A}{1 - \tan A}\), we can follow these steps:

### Step 1: Rewrite \(\tan A\) and \(\cot A\) in terms of sine and cosine
We know that:
\[ \tan A = \frac{\sin A}{\cos A} \quad \text{and} \quad \cot A = \frac{\cos A}{\sin A} \]
Substituting these into the expression gives:
\[ \frac{\frac{\sin A}{\cos A}}{1 - \frac{\cos A}{\sin A}} + \frac{\frac{\cos A}{\sin A}}{1 - \frac{\sin A}{\cos A}} \]

### Step 2: Simplify the denominators
The denominators can be simplified:
\[ 1 - \frac{\cos A}{\sin A} = \frac{\sin A - \cos A}{\sin A} \quad \text{and} \quad 1 - \frac{\sin A}{\cos A} = \frac{\cos A - \sin A}{\cos A} \]
Now substituting these back into the expression gives:
\[ \frac{\frac{\sin A}{\cos A}}{\frac{\sin A - \cos A}{\sin A}} + \frac{\frac{\cos A}{\sin A}}{\frac{\cos A - \sin A}{\cos A}} \]

### Step 3: Simplify the fractions
This can be simplified further:
\[ \frac{\sin^2 A}{\cos A (\sin A - \cos A)} + \frac{\cos^2 A}{\sin A (\cos A - \sin A)} \]
Note that \(\cos A - \sin A = -(\sin A - \cos A)\), so we can rewrite the second term:
\[ \frac{\sin^2 A}{\cos A (\sin A - \cos A)} - \frac{\cos^2 A}{\sin A (\sin A - \cos A)} \]

### Step 4: Combine the fractions
Now, we can combine the two fractions:
\[ \frac{\sin^2 A - \cos^2 A}{(\sin A - \cos A)(\sin A \cos A)} \]

### Step 5: Factor the numerator
The numerator can be factored using the difference of squares:
\[ \sin^2 A - \cos^2 A = (\sin A - \cos A)(\sin A + \cos A) \]
Substituting this back gives:
\[ \frac{(\sin A - \cos A)(\sin A + \cos A)}{(\sin A - \cos A)(\sin A \cos A)} \]

### Step 6: Cancel common terms
We can cancel \((\sin A - \cos A)\) from the numerator and denominator (assuming \(\sin A \neq \cos A\)):
\[ \frac{\sin A + \cos A}{\sin A \cos A} \]

### Final Expression
Thus, the final simplified expression is:
\[ \frac{\sin A + \cos A}{\sin A \cos A} \]