Animated Solution for Mathematics - Inverse Trigonometric Functions: Let $a,b,c$ be positive real numbers. Let $\theta ={\tan }^{-1}\sqrt{\frac{a(a+b+c)}{bc}}+{\tan }^{-1}\sqrt{\frac{b(a+b+c)}{ca}}+{\tan }^{-1}\sqrt{\frac{c(a+b+c)}{ab}}$. Then $\tan \theta =$ .........

Imagine you are staring at a massive trigonometric expression. It looks like a monster, but in JEE Advanced, whenever you see a repeating, bulky term like $(a+b+c)$, it is not there to scare you; it is there to be simplified.

Let us define $u=a+b+c$. Suddenly, the expression breathes. We are no longer looking at a mess; we are looking at a structure.

By substituting $u$, we transform the expression into:

Now, we assign variables $x,y,z$ to these square roots. Our equation becomes $\theta ={\tan }^{-1}x+{\tan }^{-1}y+{\tan }^{-1}z$. We want to combine the first two terms, but we must be careful.

Many students blindly apply the addition formula ${\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\left(\frac{x+y}{1-xy}\right)$. This is only half the story. We must check the product $xy$.

Calculating $xy$, we get:

Substituting $u=a+b+c$, we find:

Because $a,b,c>0$, this product is strictly greater than 1. This means we must use the adjusted identity:

Now, let us simplify the fraction $\frac{x+y}{1-xy}$. The numerator $x+y$ simplifies to:

The denominator $1-xy$ becomes:

When we divide these, the term $(a+b)$ cancels out with breathtaking elegance. We are left with $-\sqrt{\frac{cu}{ab}}$, which is exactly $-z$.

Our equation for $\theta$ now stands as:

Using the property ${\tan }^{-1}(-z)=-{\tan }^{-1}z$, the terms $-{\tan }^{-1}z$ and $+{\tan }^{-1}z$ annihilate each other. We are left with $\theta =\pi$.

The question asks for $\tan \theta$, and since $\tan \pi =0$, our final answer is 0. A complex, intimidating problem, reduced to zero through the sheer power of algebraic symmetry.