The value of `(sin^3theta+cos^3theta)/(sintheta+costheta)`+sin `theta` cos `theta` is:

Options

A. sin `theta` cos `theta`
B. tan `theta`
C. cot`theta`
D. 1

Answer

To solve the expression \(\frac{\sin^3 \theta + \cos^3 \theta}{\sin \theta + \cos \theta} + \sin \theta \cos \theta\), we will follow these steps:

### Step 1: Use the identity for the sum of cubes
We can use the identity for the sum of cubes, which states:
\[ A^3 + B^3 = (A + B)(A^2 - AB + B^2) \]
Here, let \(A = \sin \theta\) and \(B = \cos \theta\). Thus, we have:
\[ \sin^3 \theta + \cos^3 \theta = (\sin \theta + \cos \theta)(\sin^2 \theta - \sin \theta \cos \theta + \cos^2 \theta) \]

### Step 2: Substitute the identity into the expression
Substituting this back into our expression gives:
\[ \frac{(\sin \theta + \cos \theta)(\sin^2 \theta - \sin \theta \cos \theta + \cos^2 \theta)}{\sin \theta + \cos \theta} + \sin \theta \cos \theta \]
Since \(\sin \theta + \cos \theta\) is not zero, we can cancel it out:
\[ \sin^2 \theta - \sin \theta \cos \theta + \cos^2 \theta + \sin \theta \cos \theta \]

### Step 3: Simplify the expression
Now we simplify the remaining expression:
\[ \sin^2 \theta + \cos^2 \theta - \sin \theta \cos \theta + \sin \theta \cos \theta \]
The \(-\sin \theta \cos \theta\) and \(+\sin \theta \cos \theta\) cancel each other out, leaving us with:
\[ \sin^2 \theta + \cos^2 \theta \]

### Step 4: Apply the Pythagorean identity
Using the Pythagorean identity, we know that:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]

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