Comparison of Two Trigonometric Sets P and Q

To determine the relationship between the two sets $P$ and $Q$, we analyze the defining trigonometric equations for each set.

1. Analysis of Set $P$:
The set $P$ is defined as:
$P = \{\theta : \sin\theta - \cos\theta = \sqrt{2}\cos\theta\}$

Rearranging the equation for $P$:
$\sin\theta = (\sqrt{2} + 1)\cos\theta$

Since $\cos\theta = 0$ does not satisfy this equation (as it would imply $\sin\theta = 0$, which is impossible for the same $\theta$), we can divide both sides by $\cos\theta$:
$\tan\theta = \sqrt{2} + 1 \quad \text{--- (Equation 1)}$

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2. Analysis of Set $Q$:
The set $Q$ is defined as:
$Q = \{\theta : \sin\theta + \cos\theta = \sqrt{2}\sin\theta\}$

Rearranging the equation for $Q$:
$\cos\theta = (\sqrt{2} - 1)\sin\theta$

Dividing both sides by $\cos\theta$ (noting that $\cos\theta \neq 0$):
$\frac{\sin\theta}{\cos\theta} = \frac{1}{\sqrt{2} - 1}$

By rationalizing the denominator, we get:
$\tan\theta = \frac{1}{\sqrt{2} - 1} \times \frac{\sqrt{2} + 1}{\sqrt{2} + 1} = \frac{\sqrt{2} + 1}{2 - 1} = \sqrt{2} + 1 \quad \text{--- (Equation 2)}$

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Direct Algebraic Proof of Equivalence:
Alternatively, starting with the equation for set $P$:
$\sin\theta - \cos\theta = \sqrt{2}\cos\theta$

Multiplying both sides by the non-zero constant $(\sqrt{2} + 1)$:
$(\sqrt{2} + 1)(\sin\theta - \cos\theta) = (\sqrt{2} + 1)\sqrt{2}\cos\theta$
$\sqrt{2}\sin\theta + \sin\theta - \sqrt{2}\cos\theta - \cos\theta = 2\cos\theta + \sqrt{2}\cos\theta$

Rearranging terms:
$\sqrt{2}\sin\theta + \sin\theta - \cos\theta = 2\cos\theta$
$\sin\theta + \cos\theta = \sqrt{2}\sin\theta$

This is precisely the equation defining set $Q$. Since all algebraic operations performed are completely reversible, every solution of equation $P$ is a solution of equation $Q$ and vice versa.

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Conclusion:
Since both sets are defined by equivalent trigonometric conditions, every element of $P$ belongs to $Q$, and every element of $Q$ belongs to $P$.

Therefore,
$P = Q$

Correct Option: (D)