If `sintheta+ costheta=1`, then find the general value of `theta`.

Options

A. `n pi`
B. `(npi)/2`
C. `0`
D. none of these

Explanation

To solve the equation \( \sin \theta + \cos \theta = 1 \) and find the general value of \( \theta \), we can follow these steps:

### Step 1: Square both sides of the equation
Start with the equation:
\[ \sin \theta + \cos \theta = 1 \]
Square both sides:
\[ (\sin \theta + \cos \theta)^2 = 1^2 \]

### Step 2: Expand the left-hand side
Using the identity \( (a + b)^2 = a^2 + b^2 + 2ab \):
\[ \sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta = 1 \]

### Step 3: Use the Pythagorean identity
We know from the Pythagorean identity that:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
Substituting this into the equation gives:
\[ 1 + 2 \sin \theta \cos \theta = 1 \]

### Step 4: Simplify the equation
Subtract 1 from both sides:
\[ 2 \sin \theta \cos \theta = 0 \]

### Step 5: Use the double angle identity
Recall that \( \sin 2\theta = 2 \sin \theta \cos \theta \):
\[ \sin 2\theta = 0 \]

### Step 6: Solve for \( 2\theta \)
The sine function is zero at integer multiples of \( \pi \):
\[ 2\theta = n\pi \quad \text{where } n \in \mathbb{Z} \]

### Step 7: Solve for \( \theta \)
Divide both sides by 2:
\[ \theta = \frac{n\pi}{2} \quad \text{where } n \in \mathbb{Z} \]

### Conclusion
Thus, the general value of \( \theta \) is:
\[ \theta = \frac{n\pi}{2}, \quad n \in \mathbb{Z} \]