Animated Solution for Mathematics - Trigonometry: Let $S=\{x\in (-\frac{\pi }{2},\frac{\pi }{2}):9^{1-{\tan }^{2}x}+9^{{\tan }^{2}x}=10\}$ and $\beta ={\sum }_{x\in S}{\tan }^2(\frac{x}{3})$, then $\frac{1}{6}(\beta -14{)}^2$ is equal to

Imagine you are standing before a complex-looking equation:

At first glance, it feels like a chaotic blend of exponential growth and trigonometric oscillation. However, in the world of JEE Advanced, complexity is often just a mask for elegance.

Notice the structure of the equation. We have a term $9^{{\tan }^{2}x}$ and its reciprocal:

By substituting $t=9^{{\tan }^{2}x}$, we transform this intimidating expression into a simple, friendly quadratic equation:

Now, we are on familiar ground. Multiplying by $t$ gives us $9+t^2=10t$, or:

This is a classic quadratic. Factorizing it, we find $(t-9)(t-1)=0$, leading to $t=9$ or $t=1$.

Since $t=9^{{\tan }^{2}x}$, we have two cases: 1. $9^{{\tan }^{2}x}=9^1\Rightarrow {\tan }^{2}x=1$ 2. $9^{{\tan }^{2}x}=9^0\Rightarrow {\tan }^{2}x=0$

Within the interval $(-\frac{\pi }{2},\frac{\pi }{2})$, ${\tan }^{2}x=0$ gives $x=0$, and ${\tan }^{2}x=1$ gives $x=\pm \frac{\pi }{4}$. Our set $S$ is thus $\{0,\frac{\pi }{4},-\frac{\pi }{4}\}$.

Now, we calculate $\beta ={\sum }_{x\in S}{\tan }^2(\frac{x}{3})$. Substituting our values, we get:

Since ${\tan }^2(-\theta )={\tan }^2(\theta )$, this simplifies to $\beta =0+2{\tan }^2(\frac{\pi }{12})$. We know $\frac{\pi }{12}$ radians is $15^{\circ }$, and $\tan (15^{\circ })=2-\sqrt{3}$.

Squaring this, we get:

Thus, $\beta =2(7-4\sqrt{3})=14-8\sqrt{3}$.

We are asked to find $\frac{1}{6}(\beta -14{)}^2$. Substituting $\beta =14-8\sqrt{3}$, the expression becomes:

The $14$s cancel out, leaving $\frac{1}{6}(-8\sqrt{3}{)}^2$. Squaring $-8\sqrt{3}$ gives $64\times 3=192$.

Finally, $\frac{192}{6}=32$. We have navigated the exponential, conquered the trigonometric, and arrived at the final answer: 32.