Animated Solution for Mathematics - Trigonometry: Let $S=\{\theta \in [0,2\pi ):\tan (\pi \cos \theta )+\tan (\pi \sin \theta )=0\}$. Then ${\sum }_{\theta \in S}{\sin }^2\left(\theta +\frac{\pi }{4}\right)$ is equal to

We start with the equation:

At first glance, this looks intimidating. We have trigonometric functions nested within other trigonometric functions. However, this equation implies that:

Using the fundamental property of the tangent function, $\tan (-x)=-\tan x$, we can absorb the negative sign into the argument. The equation transforms into:

In the JEE Advanced arena, the general solution for $\tan A=\tan B$ is a concept you must have etched into your memory: $A=n\pi +B$, where $n\in \mathbb{Z}$.

Applying this to our equation, we set $A=\pi \cos \theta$ and $B=-\pi \sin \theta$. This gives us:

The $\pi$ terms cancel out beautifully. Dividing the entire equation by $\pi$, we are left with:

Now, we must be careful. Can $n$ be any integer? Consider the function $f(\theta )=\sin \theta +\cos \theta$.

We know that any expression of the form $a\sin \theta +b\cos \theta$ oscillates between $-\sqrt{a^2+b^2}$ and $\sqrt{a^2+b^2}$. For our case, $a=1$ and $b=1$, so the range is $[-\sqrt{2},\sqrt{2}]$.

Since $\sqrt{2}\approx 1.414$, the only integers that fit within this range are $n\in \{-1,0,1\}$. This is a crucial realization that constrains the infinite possibilities of $n$ down to just three distinct cases.

We are asked to find the sum of ${\sin }^2(\theta +\frac{\pi }{4})$ for all valid $\theta$. Recall the identity:

Since $\sin \theta +\cos \theta =n$, we can write:

Squaring both sides, we get:

Let us analyze our three cases for $n\in \{-1,0,1\}$ within the interval $[0,2\pi )$:

1. Case $n=1$: The equation $\sin (\theta +\frac{\pi }{4})=\frac{1}{\sqrt{2}}$ has two solutions. Each contributes $\frac{1^2}{2}=\frac{1}{2}$ to the sum. Total contribution: $2\times \frac{1}{2}=1$.

2. Case $n=-1$: The equation $\sin (\theta +\frac{\pi }{4})=-\frac{1}{\sqrt{2}}$ has two solutions. Each contributes $\frac{(-1{)}^2}{2}=\frac{1}{2}$ to the sum. Total contribution: $2\times \frac{1}{2}=1$.

3. Case $n=0$: The equation $\sin (\theta +\frac{\pi }{4})=0$ has two solutions. Each contributes $\frac{0^2}{2}=0$ to the sum. Total contribution: $2\times 0=0$.

Adding these contributions together, we find the final result: