If `tan theta=(sin alpha- cos alpha)/(sin alpha+cos alpha)`, then:

Options

A. (a) `sin alpha - cos alpha = pm sqrt(2) sin theta`
B. (b) `sin alpha + cos alpha = sqrt (2) cos theta`
C. (c) `cos 2 theta = sin 2 alpha `
D. (d) `sin 2 theta + cos 2 alpha = 0`

Answer

To solve the problem, we start with the given equation:

\[ \tan \theta = \frac{\sin \alpha - \cos \alpha}{\sin \alpha + \cos \alpha} \]

### Step 1: Square both sides
We square both sides to eliminate the tangent function:

\[ \tan^2 \theta = \frac{(\sin \alpha - \cos \alpha)^2}{(\sin \alpha + \cos \alpha)^2} \]

### Step 2: Add 1 to both sides
We know that \(1 + \tan^2 \theta = \sec^2 \theta\). Therefore, we rewrite the equation:

\[ 1 + \tan^2 \theta = 1 + \frac{(\sin \alpha - \cos \alpha)^2}{(\sin \alpha + \cos \alpha)^2} \]

### Step 3: Combine the fractions
Now, we combine the fractions on the right-hand side:

\[ 1 + \tan^2 \theta = \frac{(\sin \alpha + \cos \alpha)^2 + (\sin \alpha - \cos \alpha)^2}{(\sin \alpha + \cos \alpha)^2} \]

### Step 4: Expand the numerator
Next, we expand the numerator:

\[ (\sin \alpha + \cos \alpha)^2 = \sin^2 \alpha + 2\sin \alpha \cos \alpha + \cos^2 \alpha \]
\[ (\sin \alpha - \cos \alpha)^2 = \sin^2 \alpha - 2\sin \alpha \cos \alpha + \cos^2 \alpha \]

Adding these two expansions together:

\[ \sin^2 \alpha + 2\sin \alpha \cos \alpha + \cos^2 \alpha + \sin^2 \alpha - 2\sin \alpha \cos \alpha + \cos^2 \alpha = 2(\sin^2 \alpha + \cos^2 \alpha) = 2 \]

### Step 5: Substitute back into the equation
Now substituting back into our equation gives:

\[ 1 + \tan^2 \theta = \frac{2}{(\sin \alpha + \cos \alpha)^2} \]

### Step 6: Relate to secant squared
Since \(1 + \tan^2 \theta = \sec^2 \theta\), we have:

\[ \sec^2 \theta = \frac{2}{(\sin \alpha + \cos \alpha)^2} \]

### Step 7: Cross-multiply
Cross-multiplying gives:

\[ (\sin \alpha + \cos \alpha)^2 = 2 \cos^2 \theta \]

### Step 8: Take the square root
Taking the square root of both sides, we find:

\[ \sin \alpha + \cos \alpha = \pm \sqrt{2} \cos \theta \]

### Conclusion
From the options given in the problem, we see that:

\[ \sin \alpha + \cos \alpha = \sqrt{2} \cos \theta \]

Thus, the correct answer is:

**b) \(\sin \alpha + \cos \alpha = \sqrt{2} \cos \theta\)**