Problem 23 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

Solve for \(x\) : \(\cot ^{-1} x+\cot ^{-1}\left(n^{2}-x+1\right)=\cot ^{-1}(n-1) .\)

Step-by-Step Solution

Answer

The solutions for \(x\) are given by \(x=\frac{-[(n-1)-(n^{2}-1)]\pm \sqrt{([(n-1)-(n^{2}-1)])^2-4(-n^{2}-1)}}{2}\).

Step 1: Start by observing a key identity

One can notice that \(\cot ^{-1}(a)+\cot ^{-1}(b)=\cot ^{-1}\left(\frac{a+b}{1-ab}\right)\). This is a well-known identity that would be essential. It will be applied to the given expression.

Step 2: Application of the identity

Apply the identity to the given equation to have: \(\cot ^{-1}\left(\frac{x+n^{2}-x+1}{1-x(n^{2}-x+1)}\right)=\cot ^{-1}(n-1)\). This allows one to equate the parameters of the cotangent function.

Step 3: Equating the parameters

Now equate the parameters inside the cotangent function to write an equation without the use of inverse cotangent expressions. We have: \(\frac{x+n^{2}-x+1}{1-x(n^{2}-x+1)}=n-1\) . This equation is simpler to handle. The task now is to isolate x in this equation.

Step 4: Expanding and simplifying the equation

Expand and simplify the equation as follows: \(n^{2}+1=(n-1)-x(n^{2}-x+1)\). This equation provides a clean setup to start isolating x.

Step 5: Distributing and collecting x-terms

To solve for \(x\), distribute on the right side of the equation, and move all x-terms to one side of the equation to obtain a quadratic equation, i.e., \(x^{2}+[(n-1)-(n^{2}-1)]x-n^{2}-1=0\).

Step 6: Solving the quadratic equation

Use the quadratic formula \(x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\) to finally isolate \(x\), substitute \(a=1\), \(b=[(n-1)-(n^{2}-1)]\) and \(c=-(n^{2}+1)\) from the quadratic equation.

Key Concepts

Cotangent Inverse

Inverse trigonometric functions help us find the angles when the value of the trigonometric function is known. One such function is the inverse cotangent, denoted as \(\cot^{-1}(x)\). It is the angle whose cotangent is \(x\). This means when you apply \(\cot^{-1}\) to a value, you get an angle in radians.

The range of \(\cot^{-1}(x)\) is from \(0\) to \(\pi\) radians. In the given exercise, the function \(\cot^{-1}\) is used to express angles in terms of \(x\). A key identity in solving such problems is:

• \(\cot^{-1}(a) + \cot^{-1}(b) = \cot^{-1}\left(\frac{a+b}{1-ab}\right)\)

This identity is crucial for simplifying expressions involving sums of inverse cotangent functions. Thus, understanding \(\cot^{-1}\) is pivotal in disentangling complex equations in trigonometry.

Quadratic Equation

Quadratic equations are fundamental in algebra. They are equations of the form \(ax^2 + bx + c = 0\). Solving them typically involves finding the value of \(x\) that satisfies the equation.

In the process of solving the original exercise, we encounter a quadratic equation in the form of:

• \(x^2 + bx - c = 0\)

Here, it arises from expanding and simplifying the trigonometric equation. To solve the quadratic equation, we utilize the quadratic formula:

• \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\)

By substituting in values for \(a\), \(b\), and \(c\), you can solve for \(x\). Understanding how to derive and apply this solution helps solve many problems across mathematics.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that hold true for all values within the domain. They are incredibly useful in simplifying expressions and solving equations.

• The identity used in the cotangent problem, \(\cot^{-1}(a) + \cot^{-1}(b) = \cot^{-1}\left(\frac{a+b}{1-ab}\right)\), is known for transforming sums of inverse cotangents.
• Using this identity enables the transition from inverse trigonometric function-based equations to simpler algebraic forms, facilitating easier problem solving.

Recognizing and applying such identities is a vital skill in trigonometry. They simplify complex expressions and allow us to focus on solving simpler parts of the equation.

Problem Solving in Trigonometry

Problem solving in trigonometry can initially seem daunting, but breaking the problem into known concepts makes it manageable. With the original exercise, the following steps were effectively used:

First, identify the inverse trigonometric function involved and the applicable trigonometric identity. The identity involving \(\cot^{-1}\) transformed the complex problem into a simpler one.

• Always seek to express complicated functions using known identities.
• Isolate variables through solving equations, often leading to polynomial or quadratic equations.

Next, convert the transformed equation into a standard form that is easier to handle, such as a quadratic equation. Applying algebraic techniques like expanding, distributing, and factoring can then solve for the unknowns.

• Always check the solution by plugging values back into the original equation to verify the answer.

The key to mastering problem-solving in trigonometry is understanding the process and practicing these steps across various problems.