`(1-sinAcosA)/(cosA(secA-cosecA)).(sin^(2)A-cos^(2)A)/(sin^(3)A+cos^(3)A)=?`

Options

A. sinA
B. cosA
C. tanA
D. cosecA

Answer

To solve the expression

\[ \frac{1 - \sin A \cos A}{\cos A (\sec A - \csc A)} \cdot \frac{\sin^2 A - \cos^2 A}{\sin^3 A + \cos^3 A} \]

we will break it down into manageable parts.

### Step 1: Simplify the Denominator
First, we simplify the denominator \(\cos A (\sec A - \csc A)\).

Recall that:
\[ \sec A = \frac{1}{\cos A} \quad \text{and} \quad \csc A = \frac{1}{\sin A} \]

Thus,
\[ \sec A - \csc A = \frac{1}{\cos A} - \frac{1}{\sin A} = \frac{\sin A - \cos A}{\sin A \cos A} \]

Now substituting this back into the denominator:
\[ \cos A (\sec A - \csc A) = \cos A \cdot \frac{\sin A - \cos A}{\sin A \cos A} = \frac{\cos A (\sin A - \cos A)}{\sin A \cos A} = \frac{\sin A - \cos A}{\sin A} \]

### Step 2: Substitute Back into the Expression
Now substitute this back into the original expression:
\[ \frac{1 - \sin A \cos A}{\frac{\sin A - \cos A}{\sin A}} \cdot \frac{\sin^2 A - \cos^2 A}{\sin^3 A + \cos^3 A} \]

This simplifies to:
\[ \frac{(1 - \sin A \cos A) \cdot \sin A}{\sin A - \cos A} \cdot \frac{\sin^2 A - \cos^2 A}{\sin^3 A + \cos^3 A} \]

### Step 3: Simplify \(\sin^2 A - \cos^2 A\) and \(\sin^3 A + \cos^3 A\)
Recall that:
\[ \sin^2 A - \cos^2 A = (\sin A - \cos A)(\sin A + \cos A) \]
And for \(\sin^3 A + \cos^3 A\):
\[ \sin^3 A + \cos^3 A = (\sin A + \cos A)(\sin^2 A - \sin A \cos A + \cos^2 A) = (\sin A + \cos A)(1 - \sin A \cos A) \]

### Step 4: Substitute and Simplify
Now substituting these identities back into the expression:
\[ \frac{(1 - \sin A \cos A) \cdot \sin A}{\sin A - \cos A} \cdot \frac{(\sin A - \cos A)(\sin A + \cos A)}{(\sin A + \cos A)(1 - \sin A \cos A)} \]

The \((1 - \sin A \cos A)\) cancels out, and we have:
\[ \frac{\sin A \cdot (\sin A - \cos A)}{\sin A - \cos A} = \sin A \]

### Final Answer
Thus, the final answer is:
\[ \sin A \]