Problem 16 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(\tan ^{2}[\pi(x+y)]+\cot ^{2}[\pi(x+y)]=1+\sqrt{\frac{2 x}{1+x^{2}}}\), where \(x, y \in R\), then least positive value of \(y\) is (A) \(\frac{5}{4}\) (B) \(\frac{1}{4}\) (C) \(\frac{3}{4}\) (D) 2

Step-by-Step Solution

Answer

The least positive value of \( y \) is \( \frac{1}{4} \).

Step 1: Simplify the Trigonometric Identity

We start by simplifying the expression \( \tan^2[\pi(x+y)] + \cot^2[\pi(x+y)] = 1 + \sqrt{\frac{2x}{1+x^2}} \). Note that \( \tan^2(\theta) + \cot^2(\theta) = \sec^2(\theta) \cdot \csc^2(\theta) \). Simplifying that, we get \( \sec^2(\theta) - 2 + \csc^2(\theta) = 1 \). This equation is satisfied when \( \tan^2[\pi(x+y)] + \cot^2[\pi(x+y)] = 2 \). Hence, \( 1 = \sqrt{\frac{2x}{1+x^2}} \).

Step 2: Solve for x in the Equation

Solving \( 1 = \sqrt{\frac{2x}{1+x^2}} \) gives us \( 1 = \frac{2x}{1+x^2} \). Cross-multiplying gives \( 1+x^2 = 2x \), leading to the quadratic equation \( x^2 - 2x + 1 = 0 \). This factors as \( (x-1)^2 = 0 \), yielding \( x = 1 \).

Step 3: Set Equation for y

Substitute \( x = 1 \) into the original condition \( \tan^2[\pi(x+y)] + \cot^2[\pi(x+y)] = 2 \). We find that \( \pi(x+y) = \frac{\pi}{4} \) or any multiple of \( \frac{\pi}{4} \). Therefore, \( x+y = \frac{1}{4} + k \), where \( k \) is an integer.

Step 4: Solve for y

From the equation \( x+y = \frac{1}{4} + k \) and knowing \( x = 1 \), substituting gives \( 1 + y = \frac{1}{4} + k \). Solving for \( y \), we get \( y = \frac{1}{4} + k - 1 \). Therefore, \( y = k - \frac{3}{4} \). To find the smallest positive value of \( y \), set \( k = 1 \), giving \( y = \frac{1}{4} \).

Key Concepts

Quadratic Equations

A quadratic equation is a second-degree polynomial equation that typically takes the form \( ax^2 + bx + c = 0 \), where \( a, b, \) and \( c \) are constants, and \( a eq 0 \). Understanding quadratic equations is crucial as they appear in various mathematical contexts.

In this exercise, we derived the quadratic equation \( x^2 - 2x + 1 = 0 \) from the given equation \( 1 = \frac{2x}{1+x^2} \). This form is known as a perfect square because it can be rewritten as \((x-1)^2 = 0\).

When a quadratic equation can be factored into the form \((px+q)(rx+s)=0\), solving the equation involves setting each factor to zero and solving for \(x\). Here, \((x-1)^2 = 0\) implies \(x-1=0\), giving the solution \(x=1\). This solution is crucial, as it leads to subsequent steps in solving the problem.

• The roots of a quadratic equation provide solutions where the expression equals zero.
• Factoring simplifies the process of finding solutions, making it a key strategy in solving equations.

Simplifying Expressions

Simplifying expressions is an essential skill in algebra, as it ensures easier manipulation and solving of equations. When simplifying, the goal is to reduce the expression to its most basic form while preserving its value and relationships.

In this exercise, part of the task involved simplifying the expression \( \tan^2[\pi(x+y)] + \cot^2[\pi(x+y)] \). Through trigonometric identities, it was simplified to equal 2, leading to the equation \( 1 = \sqrt{\frac{2x}{1+x^2}} \).

This simplification is critical as it translates a complex trigonometric equation into a form that is more tractable and leads directly to a quadratic equation. Simplifying trigonometric identities can often reveal hidden relationships between components, allowing for straightforward solutions.

• Use known identities such as \( \tan^2(\theta) + 1 = \sec^2(\theta) \) to simplify expressions.
• Reducing complexity helps to view the problem from different perspectives.

Solving Trigonometric Equations

Solving trigonometric equations involves finding angle measures that make the equation true, often using identities to simplify the trigonometric functions involved. In our exercise, we dealt with the equation involving \( \tan^2[\pi(x+y)] + \cot^2[\pi(x+y)] = 2 \).

Trigonometric equations are frequently solved by expressing them in terms of a common trigonometric function, simplifying to core identities. In this case, using known relationships allowed us to reduce the problem to an equation related to the tangent function. From there, determining \( x+y = \frac{1}{4} + k \) simplified the original trigonometric condition.

When solving for specific variables, such as \( y \) here, substituting known values into the simplified equation and determining the smallest positive integer or fractional solutions addresses many typical exam or textbook problems.

• Trigonometric problems often require converting to consistent units or angles for easier solution steps.
• Identifying which trigonometric identities are most applicable can save significant time and effort.