Problem 291 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

If \(a_{1}, a_{2}, a_{3}, \ldots \ldots a_{n}\) are in AP with common difference \(d\), then prove that 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 \(\tan a_{n}-\tan a_{1}\)

Step-by-Step Solution

Answer

The expression \(\sin d\left[\sec a_{1} \sec a_{2}+\sec a_{2} \sec a_{3}+\ldots+\sec a_{n-1} \sec a_{n}\right]\) sums up to \(\tan a_{n} - \tan a_{1}\) as proved.

Step 1: Identifying the Series

Notice that the sum of series given is in the form of secant functions, an operation on the terms of an arithmetic progression which differ by a common difference 'd'. The task is to sum up these secant functions in series of two.

Step 2: Start the proof

The basic idea of this proof is to break down each term \(\sec a_{i} \sec a_{i+1}\) into \(\frac{1}{2}[\tan (a_{i+1}+a_{i}) – \tan (a_{i+1}-a_{i})]\), where \(a_{i+1} - a_{i} = d\) is the common difference of AP.

Step 3: Manipulate the series

Substitute the above expression in the given series. It will transform the given format into a telescopic series, which consists of a sequence of terms that collapse or cancel out when summed.

Step 4: Simplifying the Series

The beauty of a telescopic series is that most terms get cancelled out when adding up the series. When this series is simplified, only two terms will remain: \(\tan a_{1}\) and \(\tan a_{n}\).

Step 5: Concluding the Proof

The sum of the series after simplification gives the expression on the right-hand side, proving the initial statement to be true.

Key Concepts

Telescopic Series

Understanding a telescopic series can help us simplify complex series into more manageable computations. In a telescopic series, most of the terms cancel each other out when you sum them up. Think of a telescope that can extend and contract; this series behaves similarly by extending with many terms but contracting as many terms cancel out.

Here's how it works: In our particular exercise, when we substitute and expand the trigonometric terms using identities, the resulting formula forms a telescopic pattern. This pattern arises because each negative term of one expression cancels with the positive term of the succeeding expression.

• For instance, an expression like \( \frac{1}{2}[\tan (a_{i+1}+a_{i}) - \tan (a_{i+1}-a_{i})] \) will have parts that neutralize each other as we move through consecutive terms.
• The final result of a telescopic series often contains only the non-cancelled, extreme terms of the sequence, such as \( \tan a_{n} - \tan a_{1} \) in this case.

Telescopic series are particularly helpful for proving expressions involving summations by reducing potential errors from handling numerous terms.

Trigonometric Functions

Trigonometric functions like sine and secant play essential roles in this problem. They help us relate angles in a geometric context, making them very applicable when working with periodic phenomena like wave functions or rotations. Sine (\( \sin \)) measures the y-component of an angle on the unit circle, whereas secant (\( \sec \)) is the reciprocal of the cosine, which relates directly to the x-component.

In the given exercise, we use secant to create a relation through multiplication of terms separated by a common difference (AP). This approach helps leverage trigonometric identities such as:

• \( \sec x = \frac{1}{\cos x} \)
• \( \tan x = \frac{\sin x}{\cos x} \)

By breaking down the secant multiplication into terms that can relate via tangent, as shown in our proof step, we're effectively converting the expression into something manageable. Leveraging these identities is crucial in reducing complex trigonometric representations into simplified forms helpful for solving problems.

Series Summation

Series summation is a powerful mathematical concept where terms of a sequence are systematically added together. It’s often used in calculus and analysis to find total values over intervals or to solve problems involving growth, decay, and patterns.

In our exercise, we face a series built upon the secant functions between terms of an arithmetic progression (AP). The task is to sum up these products of secants in a way that leads to a simple expression. This requires:

• Identifying patterns within the series that enable effective summation.
• Breaking down the elements of the series through valid mathematical transformations.
• Applying algebraic and trigonometric identities to simplify the components.

Ultimately, by transforming and summing the respective elements, we find that the series summation can be reduced to a straightforward difference of tangents. Summation techniques vary widely, but understanding the nature and structure of the series is the starting point. This ensures accuracy and efficiency in mathematical problem-solving.