Problem 12 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If \(\mathrm{S}\) is the sum of the first 10 terms of the series \(\tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{1}{13}\right)+\tan ^{-1}\left(\frac{1}{21}\right)+\ldots\) then \(\tan (\mathrm{S})\) is equal to: (a) \(\frac{5}{6}\) (b) \(\frac{5}{11}\) (c) \(-\frac{6}{5}\) (d) \(\frac{10}{11}\)

Step-by-Step Solution

Answer

(c) \(-\frac{5}{6}\).

Step 1: Identify the Sequence Pattern

The given sequence consists of terms \( \tan^{-1}\left(\frac{1}{3}\right), \tan^{-1}\left(\frac{1}{7}\right), \tan^{-1}\left(\frac{1}{13}\right), \tan^{-1}\left(\frac{1}{21}\right), \ldots \) These terms follow a pattern where each term is of the form \( \tan^{-1}\left(\frac{1}{n(n+1)}\right) \) with succeeding values of \( n \). The first number in the denominator is obtained by summing consecutive numbers starting from 3: 2, 3, 5, 7, 9 ... etc.

Step 2: Apply the Sum of Arctangents Formula

Recognize that we can utilize the formula \( \tan^{-1} x + \tan^{-1} y = \tan^{-1} \left( \frac{x+y}{1-xy} \right) \) for sums that can be simplified, given a pattern in their values. We first notice that \( \tan^{-1}\left(\frac{1}{n(n+1)}\right) = \tan^{-1} (k_n) - \tan^{-1} (k_{n+1}) \), with \( n \) being consecutive terms.

Step 3: Calculate the Specific Terms

For each degree of the sequence, compute the terms iteratively: \( k_1 = \frac{1}{2}, \ k_2 = \frac{1}{4}, \ k_3 = \frac{1}{6} \) etc. Simplifying each using \( \tan^{-1}\left(\frac{1}{n(n+1)}\right) = \tan^{-1}(n) - \tan^{-1}(n+1) \).

Step 4: Simplify the Entire Series

Sum up the first 10 terms of this series using property of telescoping series, where subsequent terms cancel each other out: \( S = \left(\tan^{-1}(1) - \tan^{-1}(11)\right) \). Since parts of the series cancel telescopically, the series simplifies extensively to: \( S = \frac{\pi}{4} - \tan^{-1}(11) \).

Step 5: Calculate \( \tan(S) \)

Compute \( \tan(S) \) for an internal angle in terms of known angles: \( \tan(\frac{\pi}{4} - \tan^{-1}(11)) \). Using \( \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \), replace \( \tan A = 1 \) and \( \tan B = 11 \). Simplify \( \tan(S) \): \[ \tan(S) = \frac{1 - 11}{1 + 11 \times 1} = \frac{-10}{12} = -\frac{5}{6} \].

Step 6: Verify and Select the Correct Answer

Check calculation consistency with provided multiple-choice options. The subtraction and division simplify to \( -\frac{5}{6} \). Thus, the correct choice for \( \tan(S) \) within the given options is (c) \(-\frac{5}{6}\).

Key Concepts

Sum of Arctangents

The sum of arctangents is an interesting topic in trigonometry, where we deal with the inverse tangent values of a series of numbers. Arctangent is the inverse function of tangent, represented as \( \tan^{-1} \). When you look at a sequence of arctangents, like the one in the problem where terms are \( \tan^{-1}\left(\frac{1}{3}\right), \tan^{-1}\left(\frac{1}{7}\right), \tan^{-1}\left(\frac{1}{13}\right)\), etc., they can seem quite complex at first.

When we sum these kinds of series, we often look for properties or formulas that can simplify the process. One such property is that \( \tan^{-1} x + \tan^{-1} y = \tan^{-1} \left( \frac{x+y}{1-xy} \right) \). This property is very useful, especially when the terms \( x \) and \( y \) are positioned such that their sum or difference results in a simpler term.

• Finding patterns: In any arctan series, like \( \tan^{-1}\left(\frac{1}{n(n+1)}\right)\), identifying a pattern helps break down the complexity.
• Use of cancellation: If subsequent terms cancel each other, as found in telescoping series, the sum becomes manageable.

Telescoping Series

A telescoping series is a series where most terms cancel out when summed. This concept makes evaluating sums much simpler, as it strips away unnecessary complexity. In mathematics, the idea of a telescoping series is frequently employed to evaluate sums of series like the one shown in this problem.

To put it simply, we take adjacent terms of the series and notice that many neighboring terms cancel each other out. This is similar to a telescope collapsing into a much smaller form when closed. The series in this problem demonstrates beautifully how each \( \tan^{-1}(n) \) term relates to the next \( \tan^{-1}(n+1) \), resulting in widespread cancellation when summed over several terms.

The series given in the problem, after analysis, reveals that:

• Each term can be expressed as a difference between two consecutive arctan values: \( \tan^{-1}(n) - \tan^{-1}(n+1) \).
• When summed up, neighboring differences cancel out, simplifying to just \( \tan^{-1}(1) - \tan^{-1}(11) \).

Ultimately, this leaves us with a very compact remainder, easy to compute, and highlights the powerful minimalism and efficiency of a telescoping series in problem-solving.

Inverse Tangent Formula

The inverse tangent or arctangent function plays a critical role in simplifying mathematical problems that involve angles and triangular measures. The formula for the inverse tangent, \( \tan^{-1}(x) \), is particularly useful because it maps the ratio of opposite to adjacent side of a right-angled triangle directly to an angle, representing a principal value.

This concept extends to the arithmetic of angles, such as when dealing with identities like \( \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \). This identity is pivotal while calculating \( \tan(S) \), especially when \( S \) is derived from complex sum simplifications, like the one we saw with \( \tan^{-1}(1) - \tan^{-1}(11) \).

Key components of this application include:

• Understanding the principal range of \( \tan^{-1}(x) \), typically from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \).
• Using inverse tangent results to effectively break down and simplify complex trigonometric sums.

This formula's application was crucial in determining the value of \( \tan(S) \), where knowing \( \tan(\frac{\pi}{4} - \tan^{-1}(11)) \) helped arrive at the correct numerical answer of \(-\frac{5}{6}\). The beauty of the inverse tangent formula is in how it connects the dots between angles and calculus, aiding in solving intricate expressions.