Problem 88 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(f\) be a function such that \(f:(-1,1) \rightarrow\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\). Let \(f\) satisfy the equation \(f(x)+f(y)=f\left(x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}}\right)\) The function \(f(x)\) is equal to (A) \(\cos ^{-1} x\) (B) \(\sin ^{-1} x\) (C) \(\tan ^{-1} x\) (D) \(\sec ^{-1} x\)

Step-by-Step Solution

Answer

(C) \(\tan^{-1} x\)

Step 1: Analyze the Equation

The given equation is \(f(x) + f(y) = f\left(x \sqrt{1 - y^2} + y \sqrt{1 - x^2}\right)\). This equation has a form similar to properties of inverse trigonometric functions that deals with addition identities.

Step 2: Consider Known Properties

Recall that the addition formula for \(\tan^{-1}\) is: \[\tan^{-1} a + \tan^{-1} b = \tan^{-1} \left(\frac{a + b}{1 - ab}\right) \]when \(ab < 1\). The property also implies that the tangent function maps sums of angles to the function evaluated at sums.

Step 3: Test Compatibility with \(\tan^{-1}\)

Assume \(f(x) = \tan^{-1}(x)\) and substitute this in the given function equation.The addition formula property matches the equation when transformed as: \[\tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x + y}{1 - xy}\right)\,\text{which implies}\,\tan^{-1}(x \sqrt{1 - y^2} + y \sqrt{1 - x^2}) \]parallels how the tangent angle addition works.

Step 4: Verify the Range

The range of \(f(x)\) is given as \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\), which matches the range of \(\tan^{-1}\) for inputs in \((-1, 1)\), validating our assumption.

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions allow us to find the angle that corresponds to a given trigonometric ratio. Unlike regular trigonometric functions, which take an angle as input and provide a ratio, the inverse functions do the opposite. They take the ratio and return the angle. These functions are essential when you need to find angles, especially in calculus and geometry problems.

The most commonly used inverse trigonometric functions are:

• Arcsine - \( \sin^{-1}(x) \) gives the angle whose sine is \( x \).
• Arccosine - \( \cos^{-1}(x) \) gives the angle whose cosine is \( x \).
• Arctangent - \( \tan^{-1}(x) \) gives the angle whose tangent is \( x \).

Each of these functions has a specific range:

• Arcsine and Arctangent range from \(-\frac{\pi}{2}\) to \( \frac{\pi}{2} \).
• Arccosine ranges from \( 0 \) to \( \pi \).

Understanding the range is crucial, as it ensures the functions are defined for certain inputs. This also means for a given value within the permissible range, there is a unique angle that satisfies the function.

Addition Formula

The addition formulae are key identities that relate the trigonometric functions of sums of angles to the functions of individual angles. They are extremely handy in simplifying expressions and solving trigonometric equations. For inverse tangent, the addition formula is expressed as:\[\tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right)\]provided that \( ab < 1 \).

This identity proves helpful in the problem-solving process because it allows the transformation of the angles' sum into a single inverse tangent expression. For example, if two angles are small, this simplification is very beneficial.

• It helps in recognizing patterns or specific functional forms, like those that match given problem sets.
• It ensures that formulas match known identities, making it easier to validate potential solutions.

Recognizing when the addition formula can be applied is important. It often appears in calculus and trigonometry problems involving angle transformations or inversions.

Tangent Function

The tangent function is one of the primary trigonometric functions that relate the angles of a right triangle to the ratios of the lengths of the opposite and adjacent sides. The function is defined as:\[\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}\]The inverse, denoted as \( \tan^{-1}(x) \), returns the angle whose tangent is \( x \). The tangent function itself and its inverse have particular behaviors:

• The tan function has a periodic nature, repeating every \( \pi \) radians.
• It increases without bound and then starts over at intervals of \( \pi \).
• For its inverse, the range is clipped to \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \) to ensure a unique angle output, facilitating real-world applicability.

When solving problems like the presented exercise, the properties of the tangent function help determine the possible solutions. In particular, using its corresponding inverse allows for converting messy algebraic expressions into simpler angle forms. This facet is critical for effectively utilizing trigonometric identities for problem-solving.