Animated Solution for Mathematics - Inverse Trigonometric Functions: The value of $\cot \left({\sum }_{n=1}^{50}{\tan }^{-1}\left(\frac{1}{1+n+n^2}\right)\right)$ is

Unlocking the Telescoping Series

Welcome, future engineer. Today, we are going to dismantle a problem that, at first glance, looks like a nightmare of inverse trigonometry. We are asked to find the value of $\cot \left({\sum }_{n=1}^{50}{\tan }^{-1}\left(\frac{1}{1+n+n^2}\right)\right)$.

When you see a summation involving ${\tan }^{-1}$ and a fraction, I want you to stop and breathe. This is not a problem about brute-force calculation; it is a problem about recognizing a hidden structure. In the world of JEE Advanced, this is what we call a telescoping series in disguise.

The Anatomy of the Term

Let us isolate the general term of our summation, which we will call $T_n={\tan }^{-1}\left(\frac{1}{1+n+n^2}\right)$. Our goal is to transform this into a difference of two terms. If we can write $T_n$ as $f(n+1)-f(n)$, then the sum $\sum T_n$ will collapse like a folding telescope.

Recall the fundamental identity:

This is our golden key. We need to force our fraction $\frac{1}{1+n+n^2}$ to look like $\frac{x-y}{1+xy}$.

Look at the denominator: $1+n+n^2$. If we factor $n$ out of the last two terms, we get $1+n(n+1)$. This is perfect! We have identified our $x$ and $y$. Let $x=n+1$ and $y=n$.

Now, look at the numerator. We have a $1$. Can we write $1$ as $x-y$? Yes, $(n+1)-n=1$.

The Telescoping Magic

Now, our general term $T_n$ becomes ${\tan }^{-1}\left(\frac{(n+1)-n}{1+(n+1)n}\right)$. By applying our identity, this simplifies beautifully to $T_n={\tan }^{-1}(n+1)-{\tan }^{-1}(n)$.

This is the moment of truth. When we sum this from $n=1$ to $50$, we get:

Let us write out the first few terms to see the magic happen: For $n=1$: ${\tan }^{-1}(2)-{\tan }^{-1}(1)$ For $n=2$: ${\tan }^{-1}(3)-{\tan }^{-1}(2)$ For $n=3$: ${\tan }^{-1}(4)-{\tan }^{-1}(3)$ ... For $n=50$: ${\tan }^{-1}(51)-{\tan }^{-1}(50)$

Notice the diagonal cancellation? The ${\tan }^{-1}(2)$ cancels with $-{\tan }^{-1}(2)$, the ${\tan }^{-1}(3)$ cancels with $-{\tan }^{-1}(3)$, and this chain reaction continues until only the very first negative term and the very last positive term remain. We are left with $S={\tan }^{-1}(51)-{\tan }^{-1}(1)$.

The Final Trigonometric Bridge

We are almost there. The original question asks for $\cot (S)$. Let $A={\tan }^{-1}(51)$ and $B={\tan }^{-1}(1)$. Then $\tan A=51$ and $\tan B=1$.

We need to find $\cot (A-B)$. Using the cotangent difference identity:

Substituting our values, we get:

Simplifying this fraction, we arrive at the final answer: $\frac{26}{25}$.

Look at what we just achieved. We took a daunting expression and, through the power of algebraic manipulation and pattern recognition, reduced it to a simple fraction. This is the essence of JEE Advanced mathematics.

It is not about memorizing formulas; it is about seeing the underlying structure of the problem. Keep practicing, keep questioning, and most importantly, keep falling in love with the process. You are doing great.