What is the simplified value of `(tanA)/(1-cotA)+(cotA)/(1-tanA)-(2)/(sin2A)`?

Options

A. -1
B. 0
C. 1
D. 2

Answer

To simplify the expression \(\frac{\tan A}{1 - \cot A} + \frac{\cot A}{1 - \tan A} - \frac{2}{\sin 2A}\), we will follow these steps:

### Step 1: Rewrite \(\tan A\) and \(\cot A\)
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:
\[ \frac{\frac{\sin A}{\cos A}}{1 - \frac{\cos A}{\sin A}} + \frac{\frac{\cos A}{\sin A}}{1 - \frac{\sin A}{\cos A}} - \frac{2}{\sin 2A} \]

### Step 2: Simplify the denominators
The denominators can be simplified:
\[ 1 - \cot A = 1 - \frac{\cos A}{\sin A} = \frac{\sin A - \cos A}{\sin A} \]
\[ 1 - \tan A = 1 - \frac{\sin A}{\cos A} = \frac{\cos A - \sin A}{\cos A} \]
Thus, we can rewrite the expression as:
\[ \frac{\frac{\sin A}{\cos A}}{\frac{\sin A - \cos A}{\sin A}} + \frac{\frac{\cos A}{\sin A}}{\frac{\cos A - \sin A}{\cos A}} - \frac{2}{\sin 2A} \]

### Step 3: Simplify the fractions
This simplifies to:
\[ \frac{\sin^2 A}{\cos A (\sin A - \cos A)} + \frac{\cos^2 A}{\sin A (\cos A - \sin A)} - \frac{2}{\sin 2A} \]
Notice that \(\sin 2A = 2 \sin A \cos A\), so we can rewrite the last term:
\[ - \frac{2}{2 \sin A \cos A} = -\frac{1}{\sin A \cos A} \]

### Step 4: Combine the fractions
Now we can combine the first two fractions:
\[ \frac{\sin^2 A}{\cos A (\sin A - \cos A)} - \frac{\cos^2 A}{\sin A (\sin A - \cos A)} - \frac{1}{\sin A \cos A} \]
The common denominator for the first two fractions is \(\sin A \cos A (\sin A - \cos A)\):
\[ \frac{\sin^3 A - \cos^3 A}{\sin A \cos A (\sin A - \cos A)} - \frac{1}{\sin A \cos A} \]

### Step 5: Factor the numerator
Using the identity for the difference of cubes:
\[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \]
we have:
\[ \sin^3 A - \cos^3 A = (\sin A - \cos A)(\sin^2 A + \sin A \cos A + \cos^2 A) \]
Since \(\sin^2 A + \cos^2 A = 1\), we can simplify:
\[ \sin^2 A + \sin A \cos A + \cos^2 A = 1 + \sin A \cos A \]

### Step 6: Substitute back into the expression
Now substituting back gives:
\[ \frac{(\sin A - \cos A)(1 + \sin A \cos A)}{\sin A \cos A (\sin A - \cos A)} - \frac{1}{\sin A \cos A} \]
The \((\sin A - \cos A)\) terms cancel:
\[ \frac{1 + \sin A \cos A}{\sin A \cos A} - \frac{1}{\sin A \cos A} \]
This simplifies to:
\[ \frac{1 + \sin A \cos A - 1}{\sin A \cos A} = \frac{\sin A \cos A}{\sin A \cos A} = 1 \]

### Final Answer
Thus, the simplified value of the expression is:
\[ \boxed{1} \]