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

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: Understand the identity involved

The given equation is \(\tan^2[\pi(x+y)] + \cot^2[\pi(x+y)] = 1 + \sqrt{\frac{2x}{1+x^2}}\). Notice that \(\tan^2\theta + \cot^2\theta = \tan^2\theta + \frac{1}{\tan^2\theta}\). Therefore, this is equivalent to the identity \(\sec^2\theta - 2\csc^2\theta + 1 = 1\).

Step 2: Simplify the left side

We have \(\tan^2[\pi(x+y)] + \cot^2[\pi(x+y)] = \sec^2[\pi(x+y)] - 2\). We need to find a relationship with \(\cos^2\) or \(\sin^2\) terms since these relate to \(1 + \sin^2 + \sin^2\).

Step 3: Set up known trigonometric identity

Using the identity \(\tan^2\theta + \cot^2\theta = \frac{\sin^4\theta + \cos^4\theta}{\sin^2\theta \cdot \cos^2\theta}\), we equate this to the form shown earlier in the problem. Simplify using factorization and known identities.

Step 4: Consider special values for simplification

Given we equate the form to \(1 + \sqrt{\frac{2x}{1+x^2}}\), check when the RHS is potentially equal to a known trigonometric identity. Consider special trigonometric values that result in integer or known ratios.

Step 5: Solve for conditions

We simplify \(\sqrt{\frac{2x}{1+x^2}}\) and solve the resulting simpler equation \(\tan^2[\pi(x+y)] + \cot^2[\pi(x+y)] = 2\). Assuming \(\tan[\pi(x+y)] = 1\), thus:\( [\pi(x+y)] = \frac{\pi}{4}\).

Step 6: Solve for variable

Since \([\pi(x+y)] = \frac{\pi}{4}\), then solve for \(y\) yielding \(x+y=\frac{1}{4} + n\), where \(n\) is an integer. Find \(y\) that will satisfy this such that \(x+y\) relates back to our function structure, checking smallest positive values.

Key Concepts

Understanding Angle Relationships

Trigonometric identities often involve angle relationships, which are crucial in connecting the angles with their respective trigonometric values. Here, the identity \( \tan^2\theta + \cot^2\theta = \tan^2\theta + \frac{1}{\tan^2\theta} \) shows the interplay between tangent and cotangent. Simplifying this using known identities helps reveal connections to familiar angles like \( \frac{\pi}{4} \) or \( \frac{\pi}{3} \).
To make these connections, use ladder identities such as \( \sec^2\theta = 1 + \tan^2\theta \) and \( \csc^2\theta = 1 + \cot^2\theta \). These will transform our original terms into expressions that are easier to work with. By understanding these relationships, it becomes possible to substitute equivalent values making simplification easier.
This kind of substitution allows solving complex trigonometric equations by reducing them to simpler forms, which are fundamental in calculus and problem-solving at large. For example, with the identity \( \sec^2\theta - 2 = 1 \), we can find special angles when modifying expressions becomes less arduous.

Solving Trigonometric Equations

Trigonometric equations are solved by converting them into familiar forms using identity transformations to make them easier to handle. In the exercise, addressing the equation:

• \( \tan^2[\pi(x+y)] + \cot^2[\pi(x+y)] = 1 + \sqrt{\frac{2x}{1+x^2}} \)

The equation hints a relationship between \(x\) and \(y\) inside the trigonometric function.Use known identities like \( \tan^2\theta + \cot^2\theta = \frac{\sin^4\theta + \cos^4\theta}{\sin^2\theta \cdot \cos^2\theta} \) which otherwise simplifies trigonometric expressions. Substituting equivalent identities is like a key that unlocks the original equation and transforms it into a solvable puzzle.
To solve, consider special trigonometric values. When \( \tan\theta = 1 \), \( \theta = \frac{\pi}{4} \) is possible. Thus, setting \( \pi(x+y) = \frac{\pi}{4} \), fundamentally transforms the problem from a complex duplicate into a manageable integer solution.

Problem Solving in Calculus

Calculus is enriched by trigonometric identities as it often needs simplification of complex expressions involving limits and derivatives. In this exercise, calculus problem-solving comes into play when simplifying expressions like \( \sqrt{\frac{2x}{1+x^2}} \).
Here, simplify expressions using strategic substitutions and algebraic manipulation, which are common techniques used in calculus to handle real-valued functions better. Concepts such as continuity and differentiability rely strongly on clear simplification, which is why transforming trigonometric identities into simpler forms is essential. Solving calculus problems often requires testing values to spot patterns or continuous functions.
Identifying smallest positive values, like permissible values for \(y\) in \(x+y=\frac{1}{4} + n\), demonstrates mastery in calculus problem-solving. By breaking down calculus problems, the essence becomes identifying manipulable parts and applying trigonometric knowledge to come to a viable solution, showing how deeply interconnected these mathematical disciplines are.