If `sintheta - cos theta = 7/13` and `0 lt theta lt 90^(@)`,then the value of `sin theta + cos theta` is:

Options

A. `17/13`
B. `13/17`
C. `1/13`
D. `1/17`

Answer

To solve the problem where \( \sin \theta - \cos \theta = \frac{7}{13} \) and \( 0 < \theta < 90^\circ \), we want to find the value of \( \sin \theta + \cos \theta \).

### Step 1: Define Variables
Let:
\[ x = \sin \theta + \cos \theta \]
We have:
\[ \sin \theta - \cos \theta = \frac{7}{13} \quad \text{(Equation 1)} \]

### Step 2: Square Both Equations
Now, we will square both equations:
1. From Equation 1:
\[ (\sin \theta - \cos \theta)^2 = \left(\frac{7}{13}\right)^2 \]
This expands to:
\[ \sin^2 \theta - 2\sin \theta \cos \theta + \cos^2 \theta = \frac{49}{169} \]
Since \( \sin^2 \theta + \cos^2 \theta = 1 \), we can substitute:
\[ 1 - 2\sin \theta \cos \theta = \frac{49}{169} \]

2. For \( x \):
\[ (\sin \theta + \cos \theta)^2 = x^2 \]
This expands to:
\[ \sin^2 \theta + 2\sin \theta \cos \theta + \cos^2 \theta = x^2 \]
Again substituting \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ 1 + 2\sin \theta \cos \theta = x^2 \]

### Step 3: Set Up the Equations
Now we have two equations:
1. \( 1 - 2\sin \theta \cos \theta = \frac{49}{169} \)
2. \( 1 + 2\sin \theta \cos \theta = x^2 \)

### Step 4: Solve for \( \sin \theta \cos \theta \)
From the first equation, we can isolate \( 2\sin \theta \cos \theta \):
\[ 2\sin \theta \cos \theta = 1 - \frac{49}{169} \]
Calculating the right side:
\[ 1 = \frac{169}{169} \quad \Rightarrow \quad 1 - \frac{49}{169} = \frac{169 - 49}{169} = \frac{120}{169} \]
Thus:
\[ 2\sin \theta \cos \theta = \frac{120}{169} \]
Dividing by 2:
\[ \sin \theta \cos \theta = \frac{60}{169} \]

### Step 5: Substitute Back to Find \( x^2 \)
Now substituting \( 2\sin \theta \cos \theta \) into the second equation:
\[ 1 + 2\sin \theta \cos \theta = x^2 \]
Substituting \( 2\sin \theta \cos \theta = \frac{120}{169} \):
\[ 1 + \frac{120}{169} = x^2 \]
Calculating the left side:
\[ 1 = \frac{169}{169} \quad \Rightarrow \quad x^2 = \frac{169 + 120}{169} = \frac{289}{169} \]

### Step 6: Solve for \( x \)
Taking the square root:
\[ x = \sqrt{\frac{289}{169}} = \frac{\sqrt{289}}{\sqrt{169}} = \frac{17}{13} \]

### Conclusion
Thus, the value of \( \sin \theta + \cos \theta \) is:
\[ \boxed{\frac{17}{13}} \]