Animated Solution for Mathematics - Trigonometry: Find all values of $\theta$ in the interval $(-\pi /2,\pi /2)$ satisfying the equation $(1-\tan \theta )(1+\tan \theta ){\sec }^2\theta +2^{{\tan }^2\theta }=0$.

Imagine you are standing before a complex trigonometric equation:

It looks like a chaotic mess of terms, but in the world of JEE Advanced, these problems are puzzles designed to test your ability to see the underlying order. Let us embark on a journey to dismantle this equation.

Our first instinct should always be to simplify. Look at the first two terms: $(1-\tan \theta )(1+\tan \theta )$.

This is the classic difference of squares identity, $a^2-b^2=(a-b)(a+b)$. Applying this, the product collapses into $1-{\tan }^2\theta$.

Suddenly, the equation looks much cleaner:

Now, we face the ${\sec }^2\theta$ term. In trigonometry, we prioritize uniformity. Since we have ${\tan }^2\theta$ appearing elsewhere, let us convert the secant term using the Pythagorean identity: ${\sec }^2\theta =1+{\tan }^2\theta$.

Substituting this, our equation becomes:

Do you see it? We have another difference of squares! This simplifies to:

To make this even more manageable, let us use a substitution. Let $t={\tan }^2\theta$.

Because $t$ is the square of a real number, we must remember the constraint $t\ge 0$. Our equation now transforms into a purely algebraic form:

This is a transcendental equation. We cannot solve it with standard algebra, so we turn to the power of visualization. Imagine the graph of $y=2^t$ (an exponential growth curve) and $y=t^2-1$ (a parabola). We are looking for where they meet for $t\ge 0$.

By testing small integer values, we find that at $t=3$, the left side is $2^3=8$, and the right side is $3^2-1=8$. They match perfectly!

Since $t={\tan }^2\theta =3$, we find $\tan \theta =\pm \sqrt{3}$. Within the interval $(-\pi /2,\pi /2)$, this gives us:

We have conquered the monster by breaking it down, piece by piece. Keep this mindset, and no equation will ever be too daunting.