Animated Solution for Mathematics - Inverse Trigonometric Functions: The value of $\cot \left({\sum }_{n=1}^{19}{\cot }^{-1}\left(1+{\sum }_{p=1}^{n}2p\right)\right)$ is :

The Art of Peeling the Onion: Mastering Telescoping Series. Imagine you are standing before a massive, intimidating mathematical expression. It looks like a fortress, but every fortress has a weak point. In JEE Advanced problems, that weak point is almost always the structure of the series.

We begin by looking at the innermost part of our expression: ${\sum }_{p=1}^{n}2p$. This is the foundation. We can factor out the constant $2$, leaving us with the sum of the first $n$ natural numbers.

When we multiply this by $2$, the denominators vanish, and we are left with a clean, elegant $n(n+1)$. This is our first victory; we have simplified the core of the problem.

Now, let us define our general term $T_n={\cot }^{-1}(1+n(n+1))$. Working with ${\cot }^{-1}$ can be cumbersome, but we know that ${\cot }^{-1}(x)={\tan }^{-1}\left(\frac{1}{x}\right)$ for positive $x$.

Applying this, our term becomes:

We are hunting for a telescoping structure. We want to express this as a difference of two angles. Look at the denominator: $1+n(n+1)$. This is exactly the form $1+xy$ where $x=n+1$ and $y=n$.

If we write the numerator $1$ as $(n+1)-n$, we get:

Now, we invoke the identity ${\tan }^{-1}x-{\tan }^{-1}y={\tan }^{-1}\left(\frac{x-y}{1+xy}\right)$. Our term $T_n$ transforms into ${\tan }^{-1}(n+1)-{\tan }^{-1}(n)$.

This is the 'Aha!' moment. When we sum these terms from $n=1$ to $n=19$, we get:

Expanding this, we see:

Notice the diagonal cancellation. The ${\tan }^{-1}(2)$ cancels, the ${\tan }^{-1}(3)$ cancels, and so on. Only the very first negative term, $-{\tan }^{-1}(1)$, and the very last positive term, ${\tan }^{-1}(20)$, survive.

We are left with $S={\tan }^{-1}(20)-{\tan }^{-1}(1)$. Using the difference formula again:

Finally, we apply the outer cotangent function: $\cot (S)=\cot \left({\tan }^{-1}\frac{19}{21}\right)$. Since ${\tan }^{-1}x={\cot }^{-1}\frac{1}{x}$, this becomes:

We have dismantled the fortress, piece by piece. The final answer is $\frac{21}{19}$. Keep this logic in your toolkit; the ability to see the telescoping structure is what separates the good from the elite.