Problem 47 - 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

Option C: \(x = 3 \pm 2 \sqrt{2}\) makes \(1/y\) zero.

Step 1: Understand the given equation

We start with the given equation \(\tan^{-1} y = 4 \tan^{-1} x\). This implies that the inverse tangent of \(y\) is four times the inverse tangent of \(x\). Our goal is to find conditions under which \(1/y\) becomes zero, which implies \(y\) must be approaching infinity.

Step 2: Use tangent addition formula

By taking the tangent on both sides of the equation \(\tan^{-1} y = 4 \tan^{-1} x\), we get \(y = \tan(4 \tan^{-1} x)\). To compute \(\tan(4 \theta)\) for \(\theta = \tan^{-1}(x)\), we use the multiple-angle identity: \[ \tan(4\theta) = \frac{4\tan\theta - 4\tan^3\theta}{1-6\tan^2\theta+\tan^4\theta}. \]

Step 3: Substitute \(x\) into the identity

Setting \(\tan\theta = x\), we substitute \(x\) into the above expression. Thus, \(y = \frac{4x - 4x^3}{1 - 6x^2 + x^4}\). For \(1/y\) to be zero, the denominator must be zero, which implies that \(4x - 4x^3\) is very large.

Step 4: Analyze the denominator condition

For the denominator \(1 - 6x^2 + x^4 = 0\), we have the polynomial equation \(x^4 - 6x^2 + 1 = 0\). Solving this using the quadratic formula on \(z = x^2\), i.e., \(z^2 - 6z + 1 = 0\), we get \(z = 3 \pm 2\sqrt{2}\). Thus, \(x^2 = 3 \pm 2\sqrt{2}\), leading to \(x = \pm\sqrt{3 \pm 2\sqrt{2}}\).

Step 5: Match results with options

Calculate the options. \(x = \sqrt{3+2\sqrt{2}}\) and \(x = \sqrt{3-2\sqrt{2}}\) can be paired as \((\sqrt{3+2\sqrt{2}}, \sqrt{3-2\sqrt{2}})\) which gives us combinations matching option C, as \(3\pm2\sqrt{2}\).

Key Concepts

Tangent Addition Formula

The Tangent Addition Formula is a powerful identity in trigonometry that allows us to find the tangent of the sum of two angles. It states that for two angles \(\alpha\) and \(\beta\), the tangent of their sum is given by:

• \(\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}\)

This formula is particularly useful when dealing with inverse trigonometric functions like \(\tan^{-1} y = 4\tan^{-1} x\).
In such cases, you can find the tangent of a multiple of an angle by converting the expression using identities. For instance, \(\tan(4\theta)\) can be expressed using the Tangent Addition Formula repeatedly to simplify calculations involving multiple angles.
Understanding and applying this formula is crucial for solving problems where the angle has been multiplied, such as finding \(\tan(2\theta)\) or \(\tan(4\theta)\), as it's a common approach to reduce complex trigonometric expressions.

Multiple-Angle Identity

Multiple-Angle Identities are equations that express trigonometric functions of multiple angles in terms of powers of the function evaluated at a single angle. They are very useful when simplifying expressions or solving equations involving trigonometric functions.
One key identity is for \(\tan(4\theta)\), which is essential when dealing with recursive problems involving multiple angles. The expression is:

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

In the given exercise, knowing \(\tan(4\theta)\) was crucial in simplifying the equation and solving for \(x\). It allowed us to write \(y\) in terms of \(x\), and then find when \(1/y\) became zero.
By substituting and simplifying using the Multiple-Angle Identity, you can tackle complex equations more easily, resulting in clean and solved expressions.

Quadratic Formula

The Quadratic Formula is a fundamental tool for solving second-degree polynomial equations of the form \(ax^2 + bx + c = 0\). It helps find the values of \(x\) that satisfy the equation, using the formula:

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

In the given exercise, after expressing \(y\) in the form of an equation, we derived a polynomial that needed solving. By introducing \(z = x^2\), we transformed it into the quadratic equation \(z^2 - 6z + 1 = 0\).
Solving this with the Quadratic Formula allowed finding the roots \(z = 3 \pm 2\sqrt{2}\). This substitution removed complex calculations, providing a straightforward path to derive possible \(x\) values.
Understanding how to manipulate such equations and apply the formula ensures you can efficiently solve similar problems involving trigonometric functions.