Animated Solution for Mathematics - Trigonometry: If $\tan A=\frac{1}{\sqrt{x(x^2+x+1)}}$, $\tan B=\frac{\sqrt{x}}{\sqrt{x^2+x+1}}$ and $\tan C=(x^{-3}+x^{-2}+x^{-1}{)}^{\frac{1}{2}}$, $0<A,B,C<\frac{\pi }{2}$, then $A+B$ is equal to :

Imagine you are standing before a complex, tangled web of square roots and variables. At first glance, the expressions for $\tan A$, $\tan B$, and $\tan C$ seem designed to overwhelm you.

But in the world of JEE Advanced, complexity is often just a mask for a deeper, elegant simplicity. Today, we are going to peel back that mask.

The Setup

We are given:

Our mission is to find $A+B$. The constraint $0<A,B,C<\frac{\pi }{2}$ is our compass, ensuring we stay within the safe, positive territory of the first quadrant.

When you see tangents of angles and a request for their sum, your mind should immediately jump to the compound angle formula: This is our bridge.

The Algebraic Dance

Let us perform the substitution. The numerator is:

To add these, we need a common denominator. Multiplying the second term by $\frac{\sqrt{x}}{\sqrt{x}}$, we get:

Now, look at the denominator of our formula: $1-\tan A\tan B$. When we multiply $\tan A$ and $\tan B$, the $\sqrt{x}$ terms cancel out, and the square roots in the denominator merge to leave us with $\frac{1}{x^2+x+1}$.

Subtracting this from $1$ gives us:

The Revelation

Now, we combine these. We have:

Notice the magic? The $(1+x)$ terms cancel out. The term $\frac{x^2+x+1}{\sqrt{x^2+x+1}}$ simplifies to $\sqrt{x^2+x+1}$.

We are left with:

Now, look at $\tan C$. By converting the negative exponents to $\sqrt{\frac{1}{x^3}+\frac{1}{x^2}+\frac{1}{x}}$, we find the common denominator $x^3$:

We have arrived at $\tan (A+B)=\tan C$. Given our constraints, we can confidently state $A+B=C$.

You see, the complexity was just a test of your patience and your ability to trust the process. When you break down the math, the chaos always gives way to order.