Problem 8
Question
The smallest and the largest values of \(\tan ^{-1}\left(\frac{1-x}{1+x}\right), 0 \leq x \leq 1\) are (a) \(0, \pi\) (b) \(0, \frac{\pi}{4}\) (c) \(-\frac{\pi}{4}, \frac{\pi}{4}\) (d) \(\frac{\pi}{4}, \frac{\pi}{2}\)
Step-by-Step Solution
Verified Answer
The smallest and largest values of \( \tan^{-1}\left(\frac{1-x}{1+x}\right) \), for \( 0 \leq x \leq 1 \), are 0 and \( \frac{\pi}{4} \) respectively. Hence, the correct option is (b) \( 0, \frac{\pi}{4} \).
1Step 1: Simplify the Argument
The argument of the inverse tangent function \( \frac{1-x}{1+x} \) can be recognized as the right-hand side of the identity \( \tan(\frac{\pi}{4} - a) = \frac{1 - \tan(a)}{1 + \tan(a)} \). For this equality to hold, we set \( a = \tan^{-1}(x) \). Therefore, the original expression can be simplified as \( \tan^{-1}\left(\frac{1-x}{1+x}\right) = \tan^{-1}\left(\tan\left(\frac{\pi}{4} - \tan^{-1}(x)\right)\right) \). This simplifies to \( \tan^{-1}\left(\tan\left(\frac{\pi}{4} - \tan^{-1}(x)\right)\right) = \frac{\pi}{4} - \tan^{-1}(x) \).
2Step 2: Find the Maximum and Minimum Values
Next, we find the maximum and minimum values for the simplified expression \( \frac{\pi}{4} - \tan^{-1}(x) \) given the range \( 0 \leq x \leq 1 \). For \( x = 0 \), \( \tan^{-1}(0) = 0 \), hence the expression becomes \( \frac{\pi}{4} - 0 = \frac{\pi}{4} \). For \( x = 1 \), \( \tan^{-1}(1) = \frac{\pi}{4} \), hence the expression becomes \( \frac{\pi}{4} - \frac{\pi}{4} = 0 \). Thus, the smallest and largest values are 0 and \( \frac{\pi}{4} \) respectively.
Key Concepts
Tan Inverse PropertiesFinding Maximum and Minimum Values in TrigonometryTrigonometric Identities
Tan Inverse Properties
Inverse trigonometric functions allow us to work with angles when we have the value of a trigonometric ratio. The inverse tangent, denoted as \( \tan^{-1} \) or sometimes \( \arctan \), is particularly useful in a range of mathematical problems. Understanding the properties of the inverse tangent can help simplify complex expressions.
Here are some key properties:
Here are some key properties:
- The range of \( \tan^{-1} \) is \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \) (exclusive of the end values), which means it always returns an angle within this interval.
- If \( \tan(\theta) = x \), then \( \theta = \tan^{-1}(x) \), assuming \( \theta \) lies in the range stated above.
- \( \tan^{-1} \) has the property of odd symmetry, i.e., \( \tan^{-1}(-x) = -\tan^{-1}(x) \), which is useful in various analytical and graphical interpretations.
Finding Maximum and Minimum Values in Trigonometry
Trigonometric functions oscillate between specific maximum and minimum values. To determine these extremities in trigonometric expressions, we often use calculus or well-known properties of trigonometric functions and their inverses.
For example, the range of \( \tan^{-1} \) tells us that it can produce values between \( -\frac{\pi}{2} \) and \( \frac{\pi}{2} \), and in the context of the given exercise, we specifically look at its behaviour between 0 and 1. By analyzing the behaviour of the function at these endpoints, we can establish the smallest and largest values of the expression. This conceptual approach is at the core of solving trigonometric maximum and minimum value problems and is directly applied in the given solution where the inverse tangent is evaluated at the endpoints of the interval.
Moreover, in more complex scenarios, finding maxima and minima can involve taking the derivative of the function and solving for when the derivative equals zero (indicating potential extremities), a foundational principle in calculus.
For example, the range of \( \tan^{-1} \) tells us that it can produce values between \( -\frac{\pi}{2} \) and \( \frac{\pi}{2} \), and in the context of the given exercise, we specifically look at its behaviour between 0 and 1. By analyzing the behaviour of the function at these endpoints, we can establish the smallest and largest values of the expression. This conceptual approach is at the core of solving trigonometric maximum and minimum value problems and is directly applied in the given solution where the inverse tangent is evaluated at the endpoints of the interval.
Moreover, in more complex scenarios, finding maxima and minima can involve taking the derivative of the function and solving for when the derivative equals zero (indicating potential extremities), a foundational principle in calculus.
Trigonometric Identities
Trigonometric identities are equations involving trigonometric functions that hold true for all values within their domains. These identities provide powerful tools for simplifying trigonometric expressions, solving equations, and even proving other mathematical properties.
Some commonly used trigonometric identities include:
Some commonly used trigonometric identities include:
- The Pythagorean identities, such as \( \sin^2(\theta) + \cos^2(\theta) = 1 \)
- Angle sum and difference identities, such as \( \sin(\theta \pm \phi) = \sin(\theta)\cos(\phi) \pm \cos(\theta)\sin(\phi) \)
- Double angle identities, like \( \sin(2\theta) = 2\sin(\theta)\cos(\theta) \)
Other exercises in this chapter
Problem 7
The number of solutions of the equation \(\sin ^{-1}(1-x)-2 \sin ^{-1} x=\frac{\pi}{2}\) (a) 0 (b) 1 (c) 2 (d) More than two
View solution Problem 8
Find the value of \(\tan ^{-1}\left(\frac{1}{\sqrt{2}}\right)-\tan ^{-1}\left(\frac{\sqrt{5-2 \sqrt{6}}}{1+\sqrt{6}}\right)\)
View solution Problem 9
Let \(m=\sin ^{-1}\left(a^{6}+1\right)+\cos ^{-1}\left(a^{4}+1\right)-\tan ^{-1}\left(a^{2}+1\right)\), then find the image of the line \(x+y=m\) about the \(y\
View solution Problem 9
The equation \(\sin ^{-1} x-\cos ^{-1} x=\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)\) has (a) No solution (b) Unique Solution (c) Infinite No of soln (d) None
View solution