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

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

The answer is (C), for \(x = 3 \pm 2 \sqrt{2}\).

Step 1: Express Inverse Tangent in Terms of Tangent

The equation given is \( \tan^{-1} y = 4 \tan^{-1} x \). First, equate both sides using the tangent function: \( y = \tan(4 \tan^{-1} x) \).

Step 2: Using the Identity for Multiple Angles

To simplify \( y \), we use the identity for the tangent of a multiple angle: \[ \tan(4\theta) = \frac{4\tan \theta - 4\tan^3 \theta}{1 - 6 \tan^2 \theta + \tan^4 \theta} \]. Let \( \theta = \tan^{-1} x \), then \( \tan \theta = x \). Substitute to get: \( y = \frac{4x - 4x^3}{1 - 6x^2 + x^4} \).

Step 3: Set 1/y to Zero and Simplify

We need \( \frac{1}{y} = 0 \), which implies \( y \to \infty \). For \( y \to \infty \), the denominator of \( y \) should approach zero. Set \( 1 - 6x^2 + x^4 = 0 \).

Step 4: Solve the Polynomial Equation

Solve \( 1 - 6x^2 + x^4 = 0 \). Substitute \( z = x^2 \) to get \( 1 - 6z + z^2 = 0 \). Solve the quadratic equation with the quadratic formula: \( z = \frac{6 \pm \sqrt{36 - 4}}{2} = \frac{6 \pm \sqrt{32}}{2} \). Simplifying gives \( z = 3 \pm 2\sqrt{2} \).

Step 5: Find Values of x

Replace \( z = x^2 \) with the solutions \( z = 3 + 2\sqrt{2} \) and \( z = 3 - 2\sqrt{2} \). Solve for \( x \) to get \( x^2 = 3 \pm 2\sqrt{2} \), so \( x = \pm \sqrt{3 \pm 2\sqrt{2}} \). These correspond to \( x = 3 \pm 2\sqrt{2} \).

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are essential in connecting angles with their trigonometric values. They help in identifying an angle when the value of the trigonometric function is known.
For instance, the inverse tangent function, often written as \( \tan^{-1}(x) \) or \( \arctan(x) \), provides the angle whose tangent is \( x \).
The function is commonly used in solving equations where angles are not readily available from the tangent's typical output. This function extracts the angle from the tangent ratio, acting as the 'opposite' of tangent function.

When you have an equation like \( \tan^{-1}(y) = 4 \tan^{-1}(x) \), you're looking for an angle whose tangent corresponds to \( y \) and is four times the angle whose tangent is \( x \). This concept can solve equations in contexts ranging from physics to engineering where angle determination is key.

Tangent Multiple Angle Formula

The tangent multiple angle formula is a tool used to compute the tangent of multiple angles, particularly useful when trying to simplify or solve trigonometric equations.
Specifically, for \( \tan(4\theta) \), the formula becomes:
\[ \tan(4\theta) = \frac{4 \tan \theta - 4 \tan^3 \theta}{1 - 6 \tan^2 \theta + \tan^4 \theta} \]
This formula helps simplify complex trigonometric expressions, reducing higher-degree equations into a manageable format.

• It is vital for breaking down problems that involve angles multiplied by integers.
• The numerator simplifies the linear and cubic components, while the denominator accounts for quadratic and quartic parts.

When applying this identity, as in the exercise, substitute back into any required expressions to uncover simpler forms or solve for variables.
It streamlines the equation to reveal solutions or properties of the angle being studied.

Polynomial Equations

Polynomial equations involve variables and coefficients where variables appear with non-negative integer exponents. Solving such equations sometimes requires creativity, substitutions, and formulas.
In the original exercise, we simplified the expression to zero out the denominator, leading to a zero numerator or an infinite function. Consequently, we solved a polynomial of form
\[ 1 - 6x^2 + x^4 = 0 \]
By substituting \( z = x^2 \), the equation becomes a quadratic equation: \( 1 - 6z + z^2 = 0 \).

The quadratic formula then aids in solving by finding the roots of a polynomial:
\[ z = \frac{6 \pm \sqrt{32}}{2} \]
These steps simplified finding \( x \) by tracing back through \( z = x^2 \), which retraces the solutions for different variable scenarios.

• Quadratics typically result in two solutions.


• Solutions often return to the original variable format through a reverse substitution.

Understanding polynomial solutions is fundamental in trigonometric tasks where reducing complex expressions is necessary.