Problem 36 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(a_{1}, a_{2}, \ldots, a_{n}\) are in A.P. with common difference \(d \neq 0\), then sum of the series \(\sin d\left[\sec a_{1} \sec a_{2}+\sec \right.\) \(\left.a_{2} \sec a_{3}+\ldots+\sec a_{n-1} \sec a_{n}\right]\) is (A) \(\tan a_{n}-\tan a_{1}\) (B) \(\cot a_{n}-\cot a_{1}\) (C) \(\sec a_{n}-\sec a_{1}\) (D) \(\operatorname{cosec} a_{n}-\operatorname{cosec} a_{1}\)

Step-by-Step Solution

Answer

(A) \(\tan a_n - \tan a_1\).

Step 1: Understand the given expression

The given expression is \( \sin d \left[ \sec a_{1} \sec a_{2} + \sec a_{2} \sec a_{3} + \ldots + \sec a_{n-1} \sec a_{n} \right] \). We need to find the sum of this series, knowing the terms \( a_1, a_2, \ldots, a_n \) are in arithmetic progression (A.P.) with a common difference \( d eq 0 \).

Step 2: Represent the terms in the series

Since \( a_1, a_2, \ldots, a_n \) are in A.P., we have that \( a_2 = a_1 + d, \ a_3 = a_2 + d = a_1 + 2d, \) and so on. The general formula for the \( k \)-th term is \( a_k = a_1 + (k-1)d \).

Step 3: Simplify using trigonometric identities

To simplify the given expression, notice the product \( \sec a_k \sec a_{k+1} = \frac{1}{\cos a_k \cos a_{k+1}} \). When simplified using the identity \( \sin d \sec a_k \sec a_{k+1} \), this equals \( \tan a_{k+1} - \tan a_k \).

Step 4: Rewrite the series in a telescoping form

Recognize that the terms \( \tan a_{k+1} - \tan a_k \) form a telescoping series. The sum \( (\tan a_2 - \tan a_1) + (\tan a_3 - \tan a_2) + \ldots + (\tan a_n - \tan a_{n-1}) \) simplifies to \( \tan a_n - \tan a_1 \).

Step 5: Conclude the solution

After simplification, the sum of the series \( \sin d \left[ \sec a_{1} \sec a_{2} + \sec a_{2} \sec a_{3} + \ldots + \sec a_{n-1} \sec a_{n} \right] \) is \( \tan a_n - \tan a_1 \). This matches option (A).

Key Concepts

Trigonometric Identities

Trigonometric identities are essential tools in simplifying expressions and solving equations involving trigonometric functions. One of the fundamental identities used in this exercise is the secant identity:

• The secant function is defined as \ \( \sec \theta = \frac{1}{\cos \theta} \ \), which connects secant and cosine.
• The product of secant terms, \( \sec a_k \sec a_{k+1} \), can be expressed as \( \frac{1}{\cos a_k \cos a_{k+1}} \).
• Using these identities helps to simplify the given expression and identify it as a difference of tangent terms: \( \sin d \sec a_k \sec a_{k+1} = \tan a_{k+1} - \tan a_k \).

These identities are incredibly useful in converting complex trigonometric expressions into more manageable forms, as seen in how they allow part of this series to telescope.

Telescoping Series

A telescoping series is a type of series where most terms cancel out when the series is expanded. In this exercise, recognizing the telescoping nature of the series simplifies the sum drastically. Here's a breakdown of how telescoping series work:

• The expression involves a repetitive structure, \( \tan a_{k+1} - \tan a_k \), where consecutive terms cancel each other out.
• For a telescoping series, typically only the first and the last terms remain after cancellation, simplifying the summation process.
• In our scenario, the series \((\tan a_2 - \tan a_1) + (\tan a_3 - \tan a_2) + \ldots + (\tan a_n - \tan a_{n-1})\) shrinks down to \(\tan a_n - \tan a_1\).

Understanding telescoping series allows us to efficiently find the sum by focusing on these key terms that don't cancel, which is crucial in solving such kinds of problems.

Sum of Series

The concept of the sum of series involves techniques for finding the total of a collection of sequentially arranged numbers. Here, we are dealing with a specially structured trigonometric series, simplified using identified patterns like telescoping:

• The original series within the problem is \( \sin d \left[ \sec a_{1} \sec a_{2} + \sec a_{2} \sec a_{3} + \ldots + \sec a_{n-1} \sec a_{n} \right] \), optimized using telescoping series techniques.
• Recognizing the form \( \tan a_{k+1} - \tan a_k \) simplifies computation by reducing it to the non-cancelled terms.
• The resulting sum is then \( \tan a_n - \tan a_1 \), which translates back to the original trigonometric format of the series, making solving such problems more straightforward.

Mastering the sum of series, especially with patterns like telescoping, streamlines calculations, saving time and effort on similar mathematical challenges.