Let `P={theta:sintheta-costheta=sqrt2costheta}and Q={theta: sintheta+costheta=sqrt2sintheta}` be two sets. Then

Options

A. `psubQ and Q-P=phi`
B. `QcancelsubP`
C. `PcancelsubQ`
D. P=Q

Explanation

To solve the problem, we need to analyze the two sets \( P \) and \( Q \) defined by the equations involving trigonometric functions.

### Step 1: Solve for Set \( P \)

Given:
\[ P = \{ \theta : \sin \theta - \cos \theta = \sqrt{2} \cos \theta \} \]

Rearranging the equation:
\[ \sin \theta - \cos \theta - \sqrt{2} \cos \theta = 0 \]
\[ \sin \theta = (\sqrt{2} + 1) \cos \theta \]

Now, we can express this in terms of tangent:
\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \sqrt{2} + 1 \]

### Step 2: Solve for Set \( Q \)

Given:
\[ Q = \{ \theta : \sin \theta + \cos \theta = \sqrt{2} \sin \theta \} \]

Rearranging the equation:
\[ \sin \theta + \cos \theta - \sqrt{2} \sin \theta = 0 \]
\[ \cos \theta = (\sqrt{2} - 1) \sin \theta \]

Now, we can express this in terms of tangent:
\[ \tan \theta = \frac{\cos \theta}{\sin \theta} = \frac{1}{\sqrt{2} - 1} \]

### Step 3: Simplifying \( \tan \theta \) for Set \( Q \)

To simplify \( \tan \theta \):
\[ \tan \theta = \frac{1}{\sqrt{2} - 1} \]

Rationalizing the denominator:
\[ \tan \theta = \frac{1 \cdot (\sqrt{2} + 1)}{(\sqrt{2} - 1)(\sqrt{2} + 1)} = \frac{\sqrt{2} + 1}{2 - 1} = \sqrt{2} + 1 \]

### Conclusion

From both sets \( P \) and \( Q \), we have:
\[ \tan \theta = \sqrt{2} + 1 \quad \text{for both sets} \]

Thus, we can conclude that the sets \( P \) and \( Q \) are equal:
\[ P = Q \]

### Final Answer

The correct option is that \( P \) and \( Q \) are the same.

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