Problem 17 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The value of \(\cot \left(\sum_{n=1}^{19} \cot ^{-1}\left(1+\sum_{p=1}^{n} 2 p\right)\right)\) is: (a) \(\frac{21}{19}\) (b) \(\frac{19}{21}\) (c) \(\frac{22}{23}\) (d) \(\frac{23}{22}\)

Step-by-Step Solution

Answer

The answer is \( \frac{21}{19} \).

Step 1: Understand the Problem

We need to find the value of \( \cot \left(\sum_{n=1}^{19} \cot ^{-1}\left(1+\sum_{p=1}^{n} 2p\right)\right) \). This involves calculating a series of cotangent inverses and then applying the cotangent function to their sum.

Step 2: Simplify the Inner Sum

First, simplify \( \sum_{p=1}^{n} 2p \), which is the sum of the first \( n \) even numbers. This sum is calculated as \( 2 \sum_{p=1}^{n} p = 2\left(\frac{n(n+1)}{2}\right) = n(n+1) \). Thus, \( 1 + \sum_{p=1}^{n} 2p = 1 + n(n+1) = n(n+1) + 1 \).

Step 3: Express the Summation Expression

The term \( \sum_{n=1}^{19} \cot^{-1}\left(n(n+1) + 1\right) \) is an arithmetic sum of inverse cotangent functions. By recognizing a telescoping property, we can express it as \( \cot^{-1}(1) - \cot^{-1}(20) \) due to properties of inverse trigonometric functions.

Step 4: Calculate the Cotangent

Evaluate \( \cot\left(\cot^{-1}(1) - \cot^{-1}(20)\right) \). Using the cotangent subtraction identity, we have:\[ \cot(A - B) = \frac{\cot A \cot B + 1}{\cot B - \cot A} \]Substitute \( A = \cot^{-1}(1) \) (whose cotangent is 1) and \( B = \cot^{-1}(20) \) (whose cotangent is 20):\[ \cot\left(\cot^{-1}(1) - \cot^{-1}(20)\right) = \frac{1 \cdot 20 + 1}{20 - 1} = \frac{21}{19} \].

Step 5: Check the Answer

The solution matches option (a). Therefore, the answer to the given problem is \( \frac{21}{19} \).

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are vital in determining angles when the trigonometric ratio is known. They essentially reverse the roles of initial trigonometric functions and provide an angle as an output.
For example, in the exercise above, the function involved is the inverse cotangent, represented as \( \cot^{-1}(x) \). It provides an angle whose cotangent is \( x \). Knowing how to handle inverse trigonometric functions is crucial for solving complex problems as it enables expression simplification and discovery of equivalent expressions.

• The function \( \cot^{-1}(x) \) yields a principal value that lies in the range of 0 to \( \pi \), but typically only in the first and second quadrants.
• Such functions find widespread applications in geometry, calculus, and beyond.

Understanding these fundamentals will help apply inverse trig functions effectively across varied mathematical challenges.

Cotangent Function

The cotangent function is a fundamental trigonometric function defined as the reciprocal of the tangent. Specifically, in terms of a right triangle, the cotangent of an angle \( \theta \) is the ratio of the adjacent side to the opposite side, or \( \cot(\theta) = \frac{1}{\tan(\theta)} \).
In the context of the exercise, the cotangent function is used alongside its inverse to solve a summation problem. This encompasses utilizing properties to simplify expressions.

• In the identity used: \( \cot(A - B) = \frac{\cot A \cdot \cot B + 1}{\cot B - \cot A} \), the expression for \( A \) and \( B \) is essential to understand in calculating results from inverse angles.
• Real-world contexts for using cotangent include angular measurements and wave analysis.

Therefore, grasping the cotangent function's utility aids complex computation and conceptual understanding in higher mathematics.

Series and Summation

Series and summation are mathematical techniques for the sequence and summing of terms. In trigonometry, these concepts are often used for summing angles or functions over a range. For example, in the exercise, the summation of inverse cotangent values presented a telescopic series simplification.
This specific property often helps reach a simplified result quickly by canceling intermediate terms.

• The telescopic series allows parts of a sequence to cancel out, making complex expressions more manageable.
• Understanding the arithmetic sum formula, like \( \sum_{p=1}^{n} p = \frac{n(n+1)}{2} \), is crucial for computing totals across a range efficiently.
• Applying these techniques in trigonometric contexts optimizes the method of reaching a solution path.

By developing a strong grasp of series and summation, complex problems transform into mere arithmetic exercises, enabling streamlined problem-solving.