In `DeltaABC,if cos A+ sin A -(2)/(cos B + sin B) =0,` then `(a+b)/c` is equal to

Options

A. `sqrt2`
B. 1
C. `(1)/(sqrt2)`
D. `2sqrt2`

Answer

To solve the problem, we start with the equation given in the triangle \( \Delta ABC \):

\[ \cos A + \sin A - \frac{2}{\cos B + \sin B} = 0 \]

### Step 1: Rearranging the Equation
First, we can rearrange the equation to isolate the terms involving angles A and B:

\[ \cos A + \sin A = \frac{2}{\cos B + \sin B} \]

### Step 2: Cross Multiplying
Next, we cross-multiply to eliminate the fraction:

\[ (\cos A + \sin A)(\cos B + \sin B) = 2 \]

### Step 3: Expanding the Left Side
Now, we expand the left side of the equation:

\[ \cos A \cos B + \cos A \sin B + \sin A \cos B + \sin A \sin B = 2 \]

### Step 4: Recognizing Trigonometric Identities
We can recognize that the left-hand side can be rewritten using trigonometric identities:

\[ \cos A \cos B + \sin A \sin B = \cos(A - B) \]
\[ \cos A \sin B + \sin A \cos B = \sin(A + B) \]

Thus, we have:

\[ \cos(A - B) + \sin(A + B) = 2 \]

### Step 5: Analyzing the Maximum Values
The maximum value of \( \cos \theta \) is 1 and the maximum value of \( \sin \theta \) is also 1. Therefore, for the equation to hold true, both terms must equal their maximum values:

\[ \cos(A - B) = 1 \quad \text{and} \quad \sin(A + B) = 1 \]

### Step 6: Solving for Angles
From \( \cos(A - B) = 1 \), we conclude:

\[ A - B = 0 \implies A = B \]

From \( \sin(A + B) = 1 \), we have:

\[ A + B = \frac{\pi}{2} \]

### Step 7: Finding Angle C
Since \( A + B + C = \pi \) in a triangle, we can find angle C:

\[ C = \pi - (A + B) = \pi - \frac{\pi}{2} = \frac{\pi}{2} \]

### Step 8: Applying the Sine Rule
Now, we apply the sine rule which states:

\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k \]

Since \( A = B \) and \( C = \frac{\pi}{2} \):

\[ \sin A = \sin B = \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}, \quad \sin C = \sin\left(\frac{\pi}{2}\right) = 1 \]

### Step 9: Expressing Sides in Terms of k
Now we can express the sides in terms of \( k \):

\[ a = k \cdot \frac{1}{\sqrt{2}}, \quad b = k \cdot \frac{1}{\sqrt{2}}, \quad c = k \]

### Step 10: Finding \(\frac{a + b}{c}\)
Finally, we need to find \( \frac{a + b}{c} \):

\[ \frac{a + b}{c} = \frac{k \cdot \frac{1}{\sqrt{2}} + k \cdot \frac{1}{\sqrt{2}}}{k} = \frac{2k \cdot \frac{1}{\sqrt{2}}}{k} = \frac{2}{\sqrt{2}} = \sqrt{2} \]

### Conclusion
Thus, the value of \( \frac{a + b}{c} \) is:

\[ \frac{a + b}{c} = \sqrt{2} \]