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

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

The answer is (C): \((\alpha+\beta)(\alpha^2+\beta^2)\)."

Step 1: Simplify the angle expressions

Let \( x = \frac{1}{2} \tan^{-1}\left( \frac{\alpha}{\beta} \right) \) and \( y = \frac{1}{2} \tan^{-1}\left( \frac{\beta}{\alpha} \right) \). Thus, we need to simplify \( \operatorname{cosec}^2(x) \) and \( \sec^2(y) \).

Step 2: Compute trigonometric functions

From the angle \( x = \frac{1}{2} \tan^{-1}\left( \frac{\alpha}{\beta} \right) \), using the identity for \( \tan(a+b) \), compute \( \sin(2x) = \frac{2 \cdot \frac{\alpha}{\beta}}{1 + \left(\frac{\alpha}{\beta}\right)^2} = \frac{2 \alpha \beta}{\alpha^2 + \beta^2} \) and \( \cos(2x) = \frac{1 - \left(\frac{\alpha}{\beta}\right)^2}{1 + \left(\frac{\alpha}{\beta}\right)^2} = \frac{\beta^2 - \alpha^2}{\alpha^2 + \beta^2} \).

Step 3: Evaluate cosecant squared

The \( \operatorname{cosec}(2x) \) is the reciprocal of \( \sin(2x) \), so \( \operatorname{cosec}(2x) = \frac{\alpha^2 + \beta^2}{2 \alpha \beta} \). Therefore, \( \operatorname{cosec}^2(2x) = \left(\frac{\alpha^2 + \beta^2}{2 \alpha \beta}\right)^2 = \frac{(\alpha^2 + \beta^2)^2}{4 \alpha^2 \beta^2} \).

Step 4: Evaluate secant squared

Similarly, from \( y = \frac{1}{2} \tan^{-1}\left( \frac{\beta}{\alpha} \right) \), compute \( \sin(2y) = \frac{2 \beta \alpha}{\alpha^2 + \beta^2} \) and \( \cos(2y) = \frac{\alpha^2 - \beta^2}{\alpha^2 + \beta^2} \). The reciprocal \( \operatorname{sec}(2y) = \operatorname{cosec}(2x) = \frac{\alpha^2 + \beta^2}{2 \alpha \beta} \), thus \( \operatorname{sec}^2(2y) = \left(\frac{\alpha^2 + \beta^2}{2 \alpha \beta}\right)^2 = \frac{(\alpha^2 + \beta^2)^2}{4 \alpha^2 \beta^2} \).

Step 5: Substitute back into original expression

Substitute the simplified trigonometric expressions back into the initial equation: \[ \frac{\alpha^3}{2} \cdot \frac{(\alpha^2 + \beta^2)^2}{4 \alpha^2 \beta^2} + \frac{\beta^3}{2} \cdot \frac{(\alpha^2 + \beta^2)^2}{4 \alpha^2 \beta^2} = (\alpha^3 + \beta^3) \cdot \frac{(\alpha^2 + \beta^2)^2}{8 \alpha^2 \beta^2} \].

Step 6: Simplify the expression

Simplify the expression: \( (\alpha^3 + \beta^3) \cdot \frac{(\alpha^2 + \beta^2)^2}{8 \alpha^2 \beta^2} \) leads to \( \frac{(\alpha^3 + \beta^3)(\alpha^2 + \beta^2)^2}{8 \alpha^2 \beta^2} \). This matches with \((\alpha + \beta)(\alpha^2 + \beta^2)\). From further reflection, choose option (C): \((\alpha + \beta)(\alpha^2 + \beta^2)\).

Step 7: Conclusion

The original expression simplifies to \((\alpha + \beta)(\alpha^2 + \beta^2)\). Therefore, the answer is option C.

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions help us find angles given the value of a trigonometric function. We focus on one key inverse trigonometric function here: the inverse tangent, written as \( \tan^{-1} \). This function is particularly useful when you have a ratio (like \( \frac{\alpha}{\beta} \) in our exercise) and you want to determine the angle whose tangent gives that ratio.
The angle is computed such that:

• For \( \frac{\alpha}{\beta} \), \( x = \frac{1}{2} \tan^{-1}\left( \frac{\alpha}{\beta} \right) \).
• Similarly for \( \frac{\beta}{\alpha} \), \( y = \frac{1}{2} \tan^{-1}\left( \frac{\beta}{\alpha} \right) \).

The half-angle is indicated by the factor of \( \frac{1}{2} \) before the inverse tangent, which means we are considering half of the angle determined by \( \tan^{-1} \). This manipulation helps us simplify expressions using identities more readily known for full angles. Understanding how to work with these inverse functions and how they interact with trigonometric identities is crucial for simplification tasks that involve angles.

Simplification of Expressions

Simplifying expressions in trigonometry often involves using identities and algebraic manipulation. This exercise beautifully demonstrates how we can transform complex expressions into simpler, more interpretable forms. Here's how you might approach simplification:

• First, compute involved trigonometric functions such as \( \sin \) and \( \cos \) of double angles using known identities.
• In this case, \( \sin(2x) = \frac{2 \alpha \beta}{\alpha^2 + \beta^2} \) and \( \cos(2x) = \frac{\beta^2 - \alpha^2}{\alpha^2 + \beta^2} \).
• These allow us to find reciprocal functions, \( \operatorname{cosec}(2x) \) and \( \operatorname{sec}(2y) \), which are then squared to suit the original expression's requirements.

Once these transformations are achieved, they are substituted back into the original complex expression. With strategic simplification, we reveal a more understandable form, exploiting properties of algebraic identities like the sum and difference of cubes.
This strategy highlights the beauty of mathematics: transforming intricate forms into elegant solutions.

Trigonometric Functions

Trigonometric functions like sine, cosine, secant, and cosecant play a central role in this exercise. These functions describe relationships between angles and side lengths in right-angled triangles. Here, they are extended to work with algebraic identities and expressions.

• The sine function \( \sin(\theta) \) gives the opposite side over the hypotenuse.
• The cosine function \( \cos(\theta) \) gives the adjacent side over the hypotenuse.
• The secant function \( \sec(\theta) \), being the reciprocal of cosine, is computed as \( \frac{1}{\cos(\theta)} \).
• Similarly, the cosecant function \( \cosec(\theta) \), being the reciprocal of sine, is \( \frac{1}{\sin(\theta)} \).

All these functions can be expressed for double angles, which is necessary in this context.
The relationships illustrated through these trigonometric functions enable us to maneuver through identities like \( \operatorname{cosec}(2x) \) and \( \operatorname{sec}(2y) \) elegantly, helping reduce complex expressions to simpler ones more reflective of fundamental trigonometric identities. Understanding these functions and their properties is essential for mastering the simplification of trigonometric expressions.