Animated Solution for Mathematics - Trigonometry: Determine the smallest positive value of $x$ (in degrees) for which $\tan (x+100^{\circ })=\tan (x+50^{\circ })\tan x\tan (x-50^{\circ })$.

Unlocking the Trigonometric Puzzle

Welcome, my dear student. Today, we are going to embark on a journey through a problem that, at first glance, looks like a tangled mess of trigonometric functions.

You might look at $\tan (x+100^{\circ })=\tan (x+50^{\circ })\tan x\tan (x-50^{\circ })$ and feel a sense of dread. But I want you to take a deep breath.

In the world of JEE Advanced, intimidation is often just a mask for elegance. This problem is not a monster; it is a beautifully constructed puzzle waiting for you to find the right key.

The Analytical Setup

When we face a product of tangents, our first instinct should be to look for symmetry. Notice the terms $\tan (x+50^{\circ })$ and $\tan (x-50^{\circ })$. They are perfectly balanced around $x$.

This is our 'Spark'. We aren't going to solve this by brute force; we are going to solve it by revealing the hidden structure.

We start by converting everything into the fundamental language of trigonometry: sine and cosine. By writing $\tan \theta =\frac{\sin \theta }{\cos \theta }$, we transform the right-hand side into a ratio of products. This is where the magic begins.

The Power of Identities

Now, focus on the product $\tan (x+50^{\circ })\tan (x-50^{\circ })$. When we write this as $\frac{\sin (x+50^{\circ })\sin (x-50^{\circ })}{\cos (x+50^{\circ })\cos (x-50^{\circ })}$, we are staring at a classic identity.

Recall that $\sin (A+B)\sin (A-B)={\sin }^{2}A-{\sin }^{2}B$ and $\cos (A+B)\cos (A-B)={\cos }^{2}A-{\sin }^{2}B$. By setting $A=x$ and $B=50^{\circ }$, we collapse the product into:

This is the 'Arsenal' phase. We have reduced a complex product into a simple fraction. But we aren't done yet.

To make this manageable, we use the double-angle formulas: ${\sin }^2\theta =\frac{1-\cos 2\theta }{2}$ and ${\cos }^2\theta =\frac{1+\cos 2\theta }{2}$. By substituting these, the powers vanish, and we are left with a linear expression in terms of $\cos 2x$ and $\cos 100^{\circ }$.

The denominators, both being $2$, cancel out beautifully. This is the elegance of mathematics—when you use the right tools, the complexity simply melts away.

The Algebraic Dance

With our simplified expression, we return to the original equation:

Now, we cross-multiply. I know, it looks like it will create a massive, unmanageable expression. But trust the process.

When you expand $\sin (x+100^{\circ })\cos x(\cos 100^{\circ }+\cos 2x)=\cos (x+100^{\circ })\sin x(\cos 100^{\circ }-\cos 2x)$, you are setting the stage for the final act. Group the terms containing $\cos 100^{\circ }$ and $\cos 2x$.

You will see the compound angle formulas $\sin (A\pm B)$ emerge from the shadows. The expression collapses into:

The Grand Finale

We are almost there. To solve this, we multiply by $2$ to invoke the product-to-sum formulas. This gives us $\sin 200^{\circ }+\sin (4x+100^{\circ })+\sin 100^{\circ }=0$.

Combining $\sin 200^{\circ }+\sin 100^{\circ }$ using the sum-to-product identity yields $\sin 40^{\circ }$. Our equation is now a simple $\sin (4x+100^{\circ })=-\sin 40^{\circ }$.

By finding the general solution and testing for the smallest positive $x$, we arrive at $x=30^{\circ }$. It is a clean, perfect integer.

This is the reward for your patience and precision. You didn't just solve a problem; you navigated a complex landscape and found the path of least resistance. Keep this mindset, and no JEE problem will ever be too daunting for you.