Animated Solution for Mathematics - Complex Numbers: Let $A=\left\{\frac{1967+1686i\sin \theta }{7-3i\cos \theta }:\theta \in \mathbb{R}\right\}$. If $A$ contains exactly one positive integer $n$, then the value of $n$ is

Unlocking the Complex Set $A$

Imagine you are standing before a complex, intimidating expression:

At first glance, it looks like a chaotic mess of trigonometric functions and imaginary units. But in the world of JEE Advanced, such expressions are rarely as chaotic as they appear. They are puzzles, and every puzzle has a key. Our goal is to find the unique positive integer $n$ that lives within this set $A$.

The Realization of Reality

The problem asks for an integer $n$. As we discussed, an integer is, by definition, a purely real number.

If our complex number $z$ is to be an integer, it must have no imaginary component. This is our first, most powerful insight: $\text{Im}(z)=0$.

This condition is the lighthouse guiding us through the fog of the algebra. We do not need to solve for $z$ in its entirety; we only need to ensure that the imaginary part vanishes.

The Rationalization Strategy

To isolate the imaginary part, we must clear the complex number from the denominator. We use the classic tool of rationalization by multiplying the numerator and the denominator by the conjugate of the denominator: $7+3i\cos \theta$.

The denominator becomes $(7-3i\cos \theta )(7+3i\cos \theta )$, which simplifies beautifully using the difference of squares identity, $(a-b)(a+b)=a^2-b^2$.

Since $i^2=-1$, this becomes $49+9{\cos }^2\theta$. This is a purely real, positive denominator—a massive relief!

The Trigonometric Bridge

Now, we turn to the numerator. We only care about the imaginary part. By multiplying $(1967+1686i\sin \theta )$ by $(7+3i\cos \theta )$, we extract the imaginary terms: $1967\times 3\cos \theta$ and $1686\sin \theta \times 7$.

Setting this sum to zero gives us:

With a bit of algebraic manipulation, we find that $11802\sin \theta =-5901\cos \theta$. Dividing by $11802\cos \theta$, we arrive at the elegant result:

The Elegant Cancellation

We are almost there. We need the real part of $z$ to find $n$. Here is where we use the observation that $1967=281\times 7$ and $1686=281\times 6$.

Factoring out $281$ transforms our expression into:

When we multiply by the conjugate again to find the real part, we get:

Using the double angle identity $2\sin \theta \cos \theta =\sin 2\theta$, we simplify the numerator to $281(49-9\sin 2\theta )$.

Substituting $\tan \theta =-\frac{1}{2}$ into the identities for $\sin 2\theta$ and ${\cos }^2\theta$, we find $\sin 2\theta =-\frac{4}{5}$ and ${\cos }^2\theta =\frac{4}{5}$.

When we plug these into our expression for $n$, the numerator and denominator become identical. They cancel out perfectly, leaving us with the final, triumphant answer:

Mathematics is not just about calculation; it is about finding the hidden order within the chaos. You have just mastered that art.