Problem 26 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(\tan ^{-1} y=4 \tan ^{-1} x\), then \(1 / y\) is zero for (A) \(x=1 \pm \sqrt{2}\) (B) \(x=\sqrt{2} \pm \sqrt{3}\) (C) \(x=3 \pm 2 \sqrt{2}\) (D) all values of \(x\)

Step-by-Step Solution

Answer

None of the options lead to \(1/y = 0\). All given options are incorrect.

Step 1: Understanding the Equation

We start with the equation \(\tan^{-1} y = 4 \tan^{-1} x\). This implies that the arctangent of \(y\) equals four times the arctangent of \(x\).

Step 2: Using Arctangent Identity

We use the identity \(\tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b}\) to express \(\tan 4\tan^{-1} x\) as a single tangent term. The identity \(\tan 4\theta = \frac{4\tan \theta - 4\tan^3 \theta}{1 - 6\tan^2 \theta + \tan^4 \theta}\) can be utilized here.

Step 3: Substituting into the Identity and Simplifying

By setting \(\theta = \tan^{-1} x\), we combine it in the identity for \(\tan 4\theta\):\[y = \frac{4x - 4x^3}{1 - 6x^2 + x^4}\]This equation relates \(y\) and \(x\).

Step 4: Finding \(1/y\) and Setting it to Zero

We find \(1/y\) from the previous expression:\[\frac{1}{y} = \frac{1 - 6x^2 + x^4}{4x - 4x^3} = 0\]Setting this fraction to zero implies the denominator must be \(0\):\[4x - 4x^3 = 0\]

Step 5: Solving the Equation for \(x\)

Factor the equation obtained:\[4x(1 - x^2) = 0\]Solving gives \(x = 0\) or \(1 - x^2 = 0\), leading to \(x = \pm 1\).

Step 6: Verifying Options

Check each option if it satisfies \(x = \pm 1\):- \(x = 1 \pm \sqrt{2} \) does not satisfy- \(x = \sqrt{2} \pm \sqrt{3} \) does not satisfy- \(x = 3 \pm 2\sqrt{2} \) does not satisfyHence, \(1/y\) is never zero for any values given in options.

Key Concepts

Arctangent Identity

The arctangent, denoted as \(\tan^{-1} x\), is the inverse of the tangent function. It provides an angle whose tangent is \(x\). When dealing with expressions like \(\tan^{-1} y = 4\tan^{-1} x\), understanding how to transform these identities is crucial.
The identity for tangent of a sum \(\tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b}\) is often used to simplify complex arctangent expressions. Particularly in this exercise, the identity to express \(\tan 4\theta\) is:

• \(\tan 4\theta = \frac{4\tan \theta - 4\tan^3 \theta}{1 - 6\tan^2 \theta + \tan^4 \theta}\)

Knowing and applying this identity allows you to express complex functions of arctangents in a much simpler form, which is key in solving trigonometric equations.

Trigonometric Equations

A trigonometric equation is one involving any trigonometric functions of an angle. To solve them, several strategies exist, including using known identities and algebraic manipulation.
In our problem, after identifying \(\tan 4\tan^{-1} x = \frac{4x - 4x^3}{1 - 6x^2 + x^4}\), we relate the trigonometric expression directly to another term \(y\). This approach is helpful when simplifying complex trigonometric equations.

• Equations can often be manipulated using identities
• Identifying relationships between variables via these identities is crucial

These methods help break down trigonometric equations into simpler forms that are easier to solve, allowing for step-by-step analysis and solution.

Factorization in Algebra

Factorization is the process of breaking down an expression into a product of simpler factors. It is a fundamental concept used in solving algebraic equations. In this problem, factorizing helps find solutions to \(\frac{1}{y} = 0\) by setting the corresponding equation \(4x - 4x^3 = 0\).
Here’s how it works:

• Factor out common terms, such as \(4x\) in \(4x(1-x^2) = 0\)
• Set each factor to zero: \(4x = 0\) results in \(x = 0\), and \(1-x^2 = 0\) gives \(x = \pm 1\)

By factorizing the equation obtained after simplifying the problem, you can find the values of \(x\) that satisfy \(1/y = 0\). This process is a powerful algebraic technique used frequently to solve equations and find solution sets.