The value of `sin^3 alpha ( 1+cot alpha) + cos^3 alpha (1+tan alpha)` is equal to

Options

A. `sin alpha - cos alpha`
B. `sin alpha + cos alpha`
C. `sin alpha cos alpha`
D. cosec `alpha sec alpha`

Answer

To solve the expression \( \sin^3 \alpha (1 + \cot \alpha) + \cos^3 \alpha (1 + \tan \alpha) \), we will break it down step by step.

### Step 1: Expand the expression
We start with the given expression:
\[ \sin^3 \alpha (1 + \cot \alpha) + \cos^3 \alpha (1 + \tan \alpha) \]
This can be expanded as:
\[ \sin^3 \alpha + \sin^3 \alpha \cot \alpha + \cos^3 \alpha + \cos^3 \alpha \tan \alpha \]

### Step 2: Substitute cotangent and tangent
Recall that:
\[ \cot \alpha = \frac{\cos \alpha}{\sin \alpha} \quad \text{and} \quad \tan \alpha = \frac{\sin \alpha}{\cos \alpha} \]
Substituting these into the expression gives:
\[ \sin^3 \alpha + \frac{\sin^3 \alpha \cos \alpha}{\sin \alpha} + \cos^3 \alpha + \frac{\cos^3 \alpha \sin \alpha}{\cos \alpha} \]
This simplifies to:
\[ \sin^3 \alpha + \sin^2 \alpha \cos \alpha + \cos^3 \alpha + \cos^2 \alpha \sin \alpha \]

### Step 3: Group the terms
Now, we can group the terms:
\[ (\sin^3 \alpha + \cos^3 \alpha) + (\sin^2 \alpha \cos \alpha + \cos^2 \alpha \sin \alpha) \]

### Step 4: Apply the sum of cubes formula
Using the formula for the sum of cubes, \( A^3 + B^3 = (A + B)(A^2 - AB + B^2) \), where \( A = \sin \alpha \) and \( B = \cos \alpha \):
\[ \sin^3 \alpha + \cos^3 \alpha = (\sin \alpha + \cos \alpha)(\sin^2 \alpha - \sin \alpha \cos \alpha + \cos^2 \alpha) \]
Since \( \sin^2 \alpha + \cos^2 \alpha = 1 \), we have:
\[ \sin^3 \alpha + \cos^3 \alpha = (\sin \alpha + \cos \alpha)(1 - \sin \alpha \cos \alpha) \]

### Step 5: Simplify the second group
The second group can be factored as:
\[ \sin^2 \alpha \cos \alpha + \cos^2 \alpha \sin \alpha = \sin \alpha \cos \alpha (\sin \alpha + \cos \alpha) \]

### Step 6: Combine the results
Now, we combine both parts:
\[ (\sin \alpha + \cos \alpha)(1 - \sin \alpha \cos \alpha) + \sin \alpha \cos \alpha (\sin \alpha + \cos \alpha) \]
Factoring out \( \sin \alpha + \cos \alpha \):
\[ (\sin \alpha + \cos \alpha) \left( (1 - \sin \alpha \cos \alpha) + \sin \alpha \cos \alpha \right) \]
This simplifies to:
\[ (\sin \alpha + \cos \alpha)(1) \]

### Final Result
Thus, the final value is:
\[ \sin \alpha + \cos \alpha \]