Animated Solution for Mathematics - Trigonometry: Let $P=\{\theta :\sin \theta -\cos \theta =\sqrt{2}\cos \theta \}$ and $Q=\{\theta :\sin \theta +\cos \theta =\sqrt{2}\sin \theta \}$ be two sets. Then

Welcome, students! Today, we are going to explore a problem that might look like a daunting task of set theory and trigonometry, but it is actually a beautiful exercise in algebraic symmetry.

We are given two sets, $P$ and $Q$, defined by trigonometric conditions. Our mission is to uncover the relationship between them. Are they distinct? Is one a subset of the other? Or are they, perhaps, identical?

The Algebraic Dance

Let us begin with Set $P$. The condition for an angle $\theta$ to reside in $P$ is given by the equation:

At first glance, this looks like a jumble of trigonometric functions. But remember, in mathematics, we love to group like terms. Let us move the $-\cos \theta$ from the left side to the right. As it crosses the equality, it changes sign, becoming positive:

Now, look at the right side. We have a common factor of $\cos \theta$. Let us factor it out:

If we divide both sides by $\cos \theta$ (assuming $\cos \theta eq 0$), we arrive at a very clean expression for the tangent of $\theta$:

This is the defining characteristic of Set $P$. Any angle whose tangent is $\sqrt{2}+1$ belongs to this set.

The Hidden Mirror

Now, let us turn our attention to Set $Q$. The condition here is:

We follow the same logic. Let us group the sine terms. We move the $\sin \theta$ from the left to the right side:

Again, we factor out the common term, which is $\sin \theta$ this time:

To find $\tan \theta$, we divide both sides by $\cos \theta$ and then by $(\sqrt{2}-1)$:

At this point, you might be tempted to say, "Wait, these look different!" But here is where the magic of algebra comes in. We have an irrational denominator, $\sqrt{2}-1$. We can rationalize it by multiplying the numerator and the denominator by the conjugate, $\sqrt{2}+1$:

Using the identity $(a-b)(a+b)=a^2-b^2$, the denominator becomes $(\sqrt{2}{)}^2-1^2$, which is $2-1=1$. The expression simplifies beautifully to:

The Unity of Truth

Look at what we have achieved! The condition for Set $P$ is $\tan \theta =\sqrt{2}+1$, and the condition for Set $Q$ is also $\tan \theta =\sqrt{2}+1$.

Because the conditions are identical, the sets themselves must be identical. Every angle that satisfies the condition for $P$ automatically satisfies the condition for $Q$, and vice versa.

Thus, we conclude that $P=Q$. It is a perfect example of how different algebraic paths can lead to the exact same geometric reality. Keep practicing, and you will start to see these symmetries everywhere!