Animated Solution for Mathematics - Inverse Trigonometric Functions: If ${\sum }_{r=1}^{50}{\tan }^{-1}\frac{1}{2r^2}=p$, then the value of $\tan p$ is :

We are tasked with evaluating the sum:

At first glance, this summation seems impossible to compute directly. However, in the context of JEE Advanced, we aim to simplify the general term $T_r={\tan }^{-1}\left(\frac{1}{2r^2}\right)$ to enable a telescoping effect.

To achieve this, we utilize the trigonometric identity:

Our goal is to manipulate the argument of the inverse tangent into the form $\frac{x-y}{1+xy}$. We begin by multiplying the numerator and denominator by $2$:

Next, we force the constant $1$ into the denominator:

Recognizing that $4r^2-1$ is a difference of squares, we factor it as $(2r-1)(2r+1)$. The numerator $2$ can be expressed as the difference $(2r+1)-(2r-1)$. Thus, the term becomes:

Applying the identity, we successfully transform $T_r$ into:

As we sum from $r=1$ to $50$, the terms cancel out like dominoes. Let us visualize the expansion:

For $r=1$: ${\tan }^{-1}(3)-{\tan }^{-1}(1)$ For $r=2$: ${\tan }^{-1}(5)-{\tan }^{-1}(3)$ For $r=3$: ${\tan }^{-1}(7)-{\tan }^{-1}(5)$

The positive part of one term cancels the negative part of the subsequent term. This chain reaction continues until only the very first negative term and the very last positive term remain:

Applying the identity once more to combine these two values:

Simplifying the fraction:

Therefore, the final result is: