Problem 21 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(\tan ^{-1} \mathrm{y}=\tan ^{-1} \mathrm{x}+\tan ^{-1}\left(\frac{2 \mathrm{x}}{1-\mathrm{x}^{2}}\right)\) where or \(|\mathrm{x}|<\frac{1}{\sqrt{3}}\). Then a value of \(\mathrm{y}\) is : (a) \(\frac{3 x-x^{3}}{1+3 x^{2}}\) (b) \(\frac{3 x+x^{3}}{1+3 x^{2}}\) (c) \(\frac{3 x-x^{3}}{1-3 x^{2}}\) (d) \(\frac{3 x+x^{3}}{1-3 x^{2}}\)

Step-by-Step Solution

Answer

Option (d): \(\frac{3x+x^3}{1-3x^2}\) is the value of \(y\).

Step 1: Identify the Problem

We need to find the value of \( y \) given \( \tan^{-1} y = \tan^{-1} x + \tan^{-1}\left(\frac{2x}{1-x^2}\right) \) and \( |x| < \frac{1}{\sqrt{3}} \). The solution involves using the tangent addition formula.

Step 2: Use the Tangent Addition Formula

The tangent addition formula is \( \tan^{-1}a + \tan^{-1}b = \tan^{-1}\left(\frac{a+b}{1-ab}\right) \) if \( 1-ab > 0 \). Here, \( a = x \) and \( b = \frac{2x}{1-x^2} \). Substituting these values, we find:

Step 3: Calculate Terms in Formula

Calculate \( a + b = x + \frac{2x}{1-x^2} = \frac{x(1-x^2) + 2x}{1-x^2} = \frac{x + 2x}{1-x^2} = \frac{3x}{1-x^2} \).

Step 4: Calculate Denominator for Tangent Addition

Calculate \( 1 - ab = 1 - x \cdot \frac{2x}{1-x^2} = 1 - \frac{2x^2}{1-x^2} = \frac{1-x^2-2x^2}{1-x^2} = \frac{1-3x^2}{1-x^2} \).

Step 5: Finalize the New Tangent Expression

Combine the results from the numerator and denominator to determine \( y \), so \( \tan^{-1} y = \tan^{-1}\left( \frac{3x}{1-x^2} \div \frac{1-3x^2}{1-x^2} \right) = \tan^{-1}\left(\frac{3x}{1-3x^2}\right) \). Thus, \( y = \frac{3x}{1-3x^2} \).

Step 6: Identify the Correct Answer

From the options provided in the problem statement, \( y = \frac{3x + x^3}{1 - 3x^2} \), which matches option (d).

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions, also known as arc functions, are mathematical functions that reverse the effect of the standard trigonometric functions such as sine, cosine, and tangent. They are essential for finding an angle where you know the value of the trigonometric function.

For instance, if you know that \( an \theta = x\), then the angle \(\theta\) can be found using the inverse tangent function, represented as \(\tan^{-1}(x)\). The result, \(\tan^{-1}(x)\), gives the angle whose tangent is \(x\).

• **Notation**: The notation for inverse tangent is \(\tan^{-1}(x)\), sometimes represented as \(\arctan(x)\).


• **Domain and Range**: Typically, \(\tan^{-1}(x)\) is defined for all real numbers \(x\), and the range is \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\).


• **Applications**: It's commonly used to solve angles in right triangles and in many fields like physics and engineering where angle measures are needed.

Inverse functions allow us to explore the relationships between the angle and its corresponding trigonometric value, expanding the functionality of trigonometry in solving equations and real-world applications.

Tangent Addition Formula

The tangent addition formula is a critical identity in trigonometry used to simplify and solve equations involving the tangent of the sum of two angles. The formula is written as:
\[\tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a+b}{1-ab}\right)\quad \text{if } 1-ab > 0.\]

This formula simplifies the calculation of angles or solving equations by combining tangents of two separate angles into one. Let's explore its components:

• **Components**: The formula involves two angles whose tangents are \(a\) and \(b\). This makes it perfect for summing angles in real applications.


• **Conditions**: Note the condition \(1-ab > 0\); if this isn't met, the formula doesn't apply as it ensures the argument inside the inverse tangent remains valid.

To apply the formula, substitute values for \(a\) and \(b\) from the problem, as in correcting or solving the inverse tangent problem given in an exercise or practical application. It provides a powerful tool to make complex angle computations manageable and more straightforward.

Algebraic Manipulation

Algebraic manipulation involves using various algebraic techniques to simplify or solve equations. In the context of trigonometric identities and functions, it often includes operations such as expansion, factorization, and simplification.

For example, when solving \(\tan^{-1} y = \tan^{-1} x + \tan^{-1}\left(\frac{2x}{1-x^2}\right)\), careful manipulation of fractions and expressions is necessary.

• **Expanding and Simplifying Expressions**: Combine terms by finding a common denominator. In this case, expressing terms like \(a + b\) or simplifying \(1 - ab\) while ensuring expressions can be simplified appropriately to avoid overcomplicating the equation.


• **Division of Complex Expressions**: Sometimes, terms involve division, as where \(\frac{3x}{1-x^2} \div \frac{1-3x^2}{1-x^2}\) is simplified to form \(\frac{3x}{1-3x^2}\).


• **Solving Equations**: Isolate the variable, as seen when applying formulas like tangent addition to find the unknown \(y\).

Understanding these algebraic techniques is essential for handling trigonometric identities effectively and accurately, especially as equations grow more intricate.