If `sintheta-costheta=(7)/(17)` then `sintheta+costheta=?`

Answer

To solve the problem, we need to find the value of \( \sin \theta + \cos \theta \) given that \( \sin \theta - \cos \theta = \frac{7}{17} \).

### Step-by-Step Solution:

1. **Given Equation**:
We start with the equation:
\[ \sin \theta - \cos \theta = \frac{7}{17} \]

2. **Square Both Sides**:
We square both sides of the equation:
\[ (\sin \theta - \cos \theta)^2 = \left(\frac{7}{17}\right)^2 \]
Expanding the left side using the identity \( (a - b)^2 = a^2 - 2ab + b^2 \):
\[ \sin^2 \theta - 2\sin \theta \cos \theta + \cos^2 \theta = \frac{49}{289} \]

3. **Use the Pythagorean Identity**:
We know that \( \sin^2 \theta + \cos^2 \theta = 1 \). Substitute this into the equation:
\[ 1 - 2\sin \theta \cos \theta = \frac{49}{289} \]

4. **Rearranging the Equation**:
Rearranging gives:
\[ -2\sin \theta \cos \theta = \frac{49}{289} - 1 \]
Convert 1 into a fraction with a denominator of 289:
\[ -2\sin \theta \cos \theta = \frac{49}{289} - \frac{289}{289} = \frac{49 - 289}{289} = \frac{-240}{289} \]

5. **Finding \( \sin \theta \cos \theta \)**:
Divide both sides by -2:
\[ \sin \theta \cos \theta = \frac{240}{578} = \frac{120}{289} \]

6. **Let \( P = \sin \theta + \cos \theta \)**:
We denote \( P = \sin \theta + \cos \theta \) and square it:
\[ P^2 = (\sin \theta + \cos \theta)^2 = \sin^2 \theta + 2\sin \theta \cos \theta + \cos^2 \theta \]
Again using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ P^2 = 1 + 2\sin \theta \cos \theta \]

7. **Substituting \( \sin \theta \cos \theta \)**:
Substitute \( \sin \theta \cos \theta = \frac{120}{289} \):
\[ P^2 = 1 + 2 \cdot \frac{120}{289} = 1 + \frac{240}{289} \]
Convert 1 into a fraction:
\[ P^2 = \frac{289}{289} + \frac{240}{289} = \frac{529}{289} \]

8. **Taking the Square Root**:
Finally, take the square root to find \( P \):
\[ P = \sqrt{\frac{529}{289}} = \frac{23}{17} \quad \text{or} \quad P = -\frac{23}{17} \]

### Final Answer:
Thus, \( \sin \theta + \cos \theta = \frac{23}{17} \) or \( -\frac{23}{17} \).