If `tantheta=(sinalpha-cosalpha)/(sinalpha+cosalpha)`, then `sinalpha+cosalphaandsinalpha-cosalpha` are equal to

Options

A. `sqrt(2)costheta,sqrt(2)sintheta`
B. `sqrt(2)sintheta,sqrt(2)costheta`
C. `sqrt(2)sintheta,sqrt(2)sintheta`
D. `sqrt(2)costheta,sqrt(2)costheta`

Answer

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

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

We need to find the values of \(\sin \alpha + \cos \alpha\) and \(\sin \alpha - \cos \alpha\).

### Step 1: Rewrite the equation in terms of tangent
From the equation, we can express it as:
\[ \tan \theta = \frac{A - B}{A + B} \]
where \(A = \sin \alpha\) and \(B = \cos \alpha\).

### Step 2: Use the tangent subtraction formula
We know that:
\[ \tan(\alpha - \frac{\pi}{4}) = \frac{\tan \alpha - \tan \frac{\pi}{4}}{1 + \tan \alpha \tan \frac{\pi}{4}} = \frac{\tan \alpha - 1}{1 + \tan \alpha} \]
Thus, we can equate:
\[ \tan \theta = \tan(\alpha - \frac{\pi}{4}) \]
This implies:
\[ \theta = \alpha - \frac{\pi}{4} + n\pi \quad (n \in \mathbb{Z}) \]

### Step 3: Solve for \(\alpha\)
From the above equation, we can express \(\alpha\) as:
\[ \alpha = \theta + \frac{\pi}{4} \]

### Step 4: Substitute \(\alpha\) into \(\sin \alpha + \cos \alpha\)
Now we can substitute \(\alpha\) back into the expressions we need to find:
\[ \sin \alpha + \cos \alpha = \sin\left(\theta + \frac{\pi}{4}\right) + \cos\left(\theta + \frac{\pi}{4}\right) \]

Using the sine and cosine addition formulas:
\[ \sin\left(\theta + \frac{\pi}{4}\right) = \sin \theta \cos \frac{\pi}{4} + \cos \theta \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}} \sin \theta + \frac{1}{\sqrt{2}} \cos \theta \]
\[ \cos\left(\theta + \frac{\pi}{4}\right) = \cos \theta \cos \frac{\pi}{4} - \sin \theta \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}} \cos \theta - \frac{1}{\sqrt{2}} \sin \theta \]

### Step 5: Combine the results
Now we combine these results:
\[ \sin \alpha + \cos \alpha = \left(\frac{1}{\sqrt{2}} \sin \theta + \frac{1}{\sqrt{2}} \cos \theta\right) + \left(\frac{1}{\sqrt{2}} \cos \theta - \frac{1}{\sqrt{2}} \sin \theta\right) \]
This simplifies to:
\[ \sin \alpha + \cos \alpha = \frac{1}{\sqrt{2}} \sin \theta + \frac{1}{\sqrt{2}} \cos \theta + \frac{1}{\sqrt{2}} \cos \theta - \frac{1}{\sqrt{2}} \sin \theta = \frac{2}{\sqrt{2}} \cos \theta = \sqrt{2} \cos \theta \]

### Step 6: Find \(\sin \alpha - \cos \alpha\)
Now we find \(\sin \alpha - \cos \alpha\):
\[ \sin \alpha - \cos \alpha = \sin\left(\theta + \frac{\pi}{4}\right) - \cos\left(\theta + \frac{\pi}{4}\right) \]
Using the same sine and cosine addition formulas:
\[ \sin \alpha - \cos \alpha = \left(\frac{1}{\sqrt{2}} \sin \theta + \frac{1}{\sqrt{2}} \cos \theta\right) - \left(\frac{1}{\sqrt{2}} \cos \theta - \frac{1}{\sqrt{2}} \sin \theta\right) \]
This simplifies to:
\[ \sin \alpha - \cos \alpha = \frac{1}{\sqrt{2}} \sin \theta + \frac{1}{\sqrt{2}} \cos \theta - \frac{1}{\sqrt{2}} \cos \theta + \frac{1}{\sqrt{2}} \sin \theta = \frac{2}{\sqrt{2}} \sin \theta = \sqrt{2} \sin \theta \]

### Final Result
Thus, we have:
\[ \sin \alpha + \cos \alpha = \sqrt{2} \cos \theta \]
\[ \sin \alpha - \cos \alpha = \sqrt{2} \sin \theta \]