Problem 65 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

\(\frac{\alpha^{3}}{2} \operatorname{cosec}^{2}\left(\frac{1}{2} \tan ^{-1} \frac{\alpha}{\beta}\right)+\frac{\beta^{3}}{2} \sec ^{2}\left(\frac{1}{2} \tan ^{-1} \frac{\beta}{\alpha}\right)\) is equal to (A) \((\alpha-\beta)\left(\alpha^{2}+\beta^{2}\right)\) (B) \((\alpha+\beta)\left(\alpha^{2}-\beta^{2}\right)\) (C) \((\alpha+\beta)\left(\alpha^{2}+\beta^{2}\right)\) (D) none of these

Step-by-Step Solution

Answer

Option (C): \((\alpha+\beta)(\alpha^{2}+\beta^{2})\)."

Step 1: Understanding the Problem

We need to simplify the given expression \( \frac{\alpha^{3}}{2} \operatorname{cosec}^{2}\left(\frac{1}{2} \tan^{-1}\frac{\alpha}{\beta}\right)+\frac{\beta^{3}}{2} \sec^{2}\left(\frac{1}{2} \tan^{-1} \frac{\beta}{\alpha}\right) \) and match it with one of the given options.

Step 2: Using the Trigonometric Identity

Recall that \( \operatorname{cosec}^{2}(2x) = \frac{1}{1 - \sin^{2}(x)} \) and \( \sec^{2}(2x) = \frac{1}{1 - \cos^{2}(x)} \). The half-angle identities \( \sin(2x) = \frac{2\sin(x)\cos(x)}{1-\cos^2(x)} \) and \( \cos(2x) = \frac{1-2\sin^2(x)}{\cos^2(x)} \) are applicable when simplifying the expression.

Step 3: Transforming the Angles

Let's substitute \( x = \tan^{-1}\frac{\alpha}{\beta} \) and \( y = \tan^{-1}\frac{\beta}{\alpha} \). This gives us \( \tan(2x) = \frac{2\tan(x)}{1-\tan^2(x)} \) and \( \tan(2y) = \frac{2\tan(y)}{1-\tan^2(y)} \). By simplifying, we get \( \operatorname{cosec}^{2}(2x) \) and \( \sec^{2}(2y) \).

Key Concepts

Half-Angle Identities

Half-angle identities are invaluable tools in trigonometry, especially when dealing with angles that are not straightforward or when simplifying complex expressions. Let's break down the main concepts:

• Half-angle identities involve trigonometric functions such as sine, cosine, tangent, cosecant, secant, and cotangent.
• They are particularly useful for angles like \( \theta/2 \). Knowing these formulas helps in converting and solving expressions involving these angles.
• For example, the half-angle identity for sine is \( \sin(x/2) = \sqrt{\frac{1 - \cos(x)}{2}} \).
• Similarly, for cosine, we have \( \cos(x/2) = \sqrt{\frac{1 + \cos(x)}{2}} \).
• These identities allow us to express trigonometric functions of half-angles in terms of the full-angle, simplifying the process of solving trigonometric equations or simplifying expressions.

In the original exercise, half-angle identities were applied to simplify expressions involving inverse trigonometric functions. By using identities such as \( \operatorname{cosec}^{2}(2x) = \frac{1}{1 - \sin^{2}(x)} \), one can simplify complex trigonometric expressions.

Simplifying Expressions

Simplifying expressions in mathematics is akin to tidying a cluttered room—it makes the expression clearer and easier to work with. Simplification involves using mathematical identities and properties to break down a complicated expression to its simplest form.

• Key simplification strategies include combining like terms, canceling out factors, using trigonometric identities, and substituting values.
• A common approach is also using half-angle identities or other trigonometric identities to rearrange the expression.
• In the exercise example, simplifying meant transforming the angles using half-angle identities and inverse trigonometric functions like \( \tan(x) \).

The goal of these simplifications is to make it possible to identify the expression's value more easily with respect to the provided options.

Inverse Trigonometric Functions

Inverse trigonometric functions are fascinating as they allow us to find angles when given trigonometric values. Let's delve into some critical aspects:

• These functions include \( \sin^{-1}, \cos^{-1}, \tan^{-1}, \text{etc.} \) and they serve to find the angle that corresponds to a specific trigonometric function value.
• For instance, if \( \tan^{-1}(x) \) is involved, it means finding the angle whose tangent is \( x \).
• Inverse trigonometric functions are crucial when dealing with problems that involve angle calculations, making them prevalent in integration, solving trigonometric equations, and angle transformations.

In the context of the original problem, inverse trigonometric functions are used to identify the angles \( x = \tan^{-1}(\frac{\alpha}{\beta}) \) and \( y = \tan^{-1}(\frac{\beta}{\alpha}) \) before applying trigonometric simplifications.
Understanding how to handle these functions is key to handling complex trigonometric equations effectively.