Animated Solution for Mathematics - Trigonometry: Let $S=\left[-\pi ,\frac{\pi }{2}\right)-\left\{-\frac{\pi }{2},-\frac{\pi }{4},\frac{3\pi }{4},\frac{\pi }{4}\right\}$. Then the number of elements in the set $A=\{\theta \in S:\tan \theta (1+\sqrt{5}\tan (2\theta ))=\sqrt{5}-\tan (2\theta )\}$ is

Imagine you are standing at the threshold of a complex trigonometric landscape. You are presented with the equation:

At first glance, it looks like a tangled mess of variables and radicals. In the world of JEE Advanced, complexity is often just a mask for elegance. Our mission is to peel back that mask.

Let us begin by expanding the left-hand side of the equation. Distributing the $\tan \theta$, we get:

Now, look at the terms. We have $\tan \theta$ and $\tan 2\theta$ on both sides. Let us group them by adding $\tan 2\theta$ to both sides and subtracting $\sqrt{5}\tan \theta \tan 2\theta$ from both sides:

This is the moment of clarity. If we factor out $\sqrt{5}$ on the right-hand side, we see:

The $\tan (3\theta )$ Breakthrough

Do you see it now? The structure below is staring us in the face:

This is the classic compound angle identity for tangent, $\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$, where $A=\theta$ and $B=2\theta$. Thus, the entire equation collapses into the beautifully simple:

We have transformed a daunting algebraic expression into a fundamental trigonometric equation. We are now looking for the intersection points of the curve $y=\tan (3\theta )$ and the horizontal line $y=\sqrt{5}$.

The General Solution

Now, we must find the general solution. Let $\tan \alpha =\sqrt{5}$, where $\alpha$ is an acute angle. The general solution for $\tan (3\theta )=\sqrt{5}$ is:

Our domain is $S=[-\pi ,\frac{\pi }{2})\setminus \{-\frac{\pi }{2},-\frac{\pi }{4},\frac{3\pi }{4},\frac{\pi }{4}\}$. We must test integer values of $n$ to find which $\theta$ values fall within this range.

For $n=-3$, $\theta =-\pi +\frac{\alpha }{3}$, which is valid. For $n=-2$, $\theta =-\frac{2\pi }{3}+\frac{\alpha }{3}$, which is also valid. For $n=-1$, $\theta =-\frac{\pi }{3}+\frac{\alpha }{3}$, which is valid.

For $n=0$, $\theta =\frac{\alpha }{3}$, which is valid. For $n=1$, $\theta =\frac{\pi }{3}+\frac{\alpha }{3}$, which is valid. If we try $n=2$, $\theta =\frac{2\pi }{3}+\frac{\alpha }{3}$, which is greater than $\frac{\pi }{2}$, so it falls outside our domain.

Ensuring Precision

Finally, we must check against the excluded values: $\{-\frac{\pi }{2},-\frac{\pi }{4},\frac{3\pi }{4},\frac{\pi }{4}\}$. Since $\tan \alpha =\sqrt{5}$, $\alpha$ is not a multiple of $\frac{\pi }{4}$ or $\frac{\pi }{2}$.

Therefore, none of our five solutions will clash with these excluded values. We have successfully identified exactly five valid elements in set $A$.

This journey shows that even the most intimidating problems can be solved with patience, the right identity, and a careful eye on the domain. Keep practicing, and you will find that the beauty of mathematics lies in these moments of perfect cancellation.