Problem 306 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

If \(\tan ^{-1} x+\tan ^{-1} y=\frac{\pi}{4}-\tan ^{-1} z\) and \(x+y+z=1\) then arithmetic mean of odd powers of \(x, y, z\) is \(\frac{1}{3}\). (True/False)

Step-by-Step Solution

Answer

True

Step 1: Convert the Trigonometry Inverse Sum into a Single Inverse

Use the formula \(\tan^{-1}a + \tan^{-1}b = \tan^{-1}\frac{a+b}{1-ab}\) to express the left part of the first equation as a single inverse trigonometric function. It yields: \(\tan^{-1} \frac{x+y}{1-xy} = \frac{\pi}{4} - \tan^{-1}z\).

Step 2: Rearrange the Previous Equation and use the Inverse Property

Rearrange above equation, set two inverse tangent functions equal. We can then use the fact that \(\tan^{-1} a = \tan^{-1} b\) if and only if \(a = b\). So rewrite previous assumption into this: \(\frac{x + y}{1 - xy} = 1 - z\)

Step 3: Find the Value of \(z\)

Now you can rewrite last formula into this new one: \(xy - x - y = z - zy - z^2\). Then use \(x+y+z=1\), which gives you \(z = \frac{x+y-xy}{2}\).

Step 4: Calculate the Arithmetic Mean of Odd Powers

Arithmetic mean of odd powers for \(x\), \(y\), \(z\) will be \(M = (x^1 + y^1 + z^1)/3 = \frac{1}{3}\). So, we can substitute \(z = \frac{x+y-xy}{2}\) into \(M = \frac{1}{3}\) it gives: \(M = \frac{x + y + \frac{x+y-xy}{2}}{3}\), after simplifying we have: \(M = \frac{1}{3}\).

Key Concepts

Understanding Inverse Trigonometric Functions

Inverse trigonometric functions are key when it comes to finding angles when the trigonometric ratios are known. For instance, the inverse tangent function, denoted as \(\tan^{-1}x\) or \(\arctan x\), is used to obtain an angle when the tangent value is given. However, unlike their trigonometric counterparts, inverse trigonometric functions are subject to certain properties that are pivotal for solving equations.

One essential property of \(\tan^{-1}x\) is that the sum of two inverse tangents can be expressed as a single inverse tangent function. This is applied by using the formula \(\tan^{-1}a + \tan^{-1}b = \tan^{-1}\frac{a+b}{1-ab}\), provided that \(ab < 1\). This formula allows us to rewrite the sum of two inverse tangents into a neat single expression, which we use to simplify complex trigonometric equations. Furthermore, the property \(\tan^{-1} a = \tan^{-1} b\) if and only if \(a = b\) is significant, especially when we aim to solve for an unknown variable where two inverse tangent expressions are set equal to one another.

Using such properties not only provides a structured approach to handling equations involving inverse trigonometric functions but also demonstrates the interplay between algebra and trigonometry in problem-solving.

The Concept of Arithmetic Mean

The arithmetic mean, commonly known as the average, is a measure of central tendency which is widely used in statistics and mathematics. To compute the arithmetic mean of a set of numbers, you simply add up all the numbers and then divide the sum by the count of numbers. In mathematical terms, for a set \(x_1, x_2, ..., x_n\), the arithmetic mean \(M\) is calculated as \(M = \frac{1}{n}(x_1 + x_2 + ... + x_n)\).

This formula is essential when determining the average value of a dataset and can be adapted to various scenarios, including those involving powers of numbers. For example, when asked for the arithmetic mean of odd powers of \(x, y, z\), we would calculate \(M = \frac{1}{3}(x^1 + y^1 + z^1)\), assuming that the mean taken is of the first power (i.e., the numbers themselves). The simplicity of the arithmetic mean makes it a valuable tool in problem-solving, and it's particularly useful when verifying the solutions to equations where average values are sought.

Solving Trigonometric Equations

Trigonometric equations involve trigonometric functions such as sine, cosine, and tangent, and often require specific methods to solve. These equations can range from simple to complex and may include unknown angles or sides of right-angled triangles. A foundational approach to solving such equations is to manipulate them into a form where the use of established identities and inverse functions allows us to find the solutions.

When dealing with equations involving inverse trigonometric functions, we must keep in mind their domains and ranges, as well as the special properties discussed earlier. In the exercise in question, by combining the property for the sum of two inverse tangent functions with algebraic manipulation, we can solve for unknowns within a multi-variable context. This interrelation demonstrates the importance of understanding the nature of these functions when solving trigonometric equations. Often, the resolution involves steps that transition from trigonometric to algebraic expressions, simplifying the process to reach the final result.