Problem 25 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(\sin x+\operatorname{cosec} x+\tan y+\cot y=4\), where \(x\) and \(y \in\left[0, \frac{\pi}{2}\right]\), then \(\tan \frac{y}{2}\) is a root of the equation (A) \(\alpha^{2}+2 \alpha+1=0\) (B) \(\alpha^{2}+2 \alpha-1=0\) (C) \(2 \alpha^{2}-2 \alpha-1=0\) (D) none of thes

Step-by-Step Solution

Answer

\( \tan \frac{y}{2} \) is a root of equation (B) \( \alpha^2 + 2\alpha - 1 = 0 \).

Step 1: Utilize trigonometric identities

We start by examining the given equation: \( \sin x + \csc x + \tan y + \cot y = 4 \). Using trigonometric identities, recall that \( \csc x = \frac{1}{\sin x} \), \( \tan y = \frac{\sin y}{\cos y} \), and \( \cot y = \frac{\cos y}{\sin y} \). These identities help simplify the equation.

Step 2: Simplify the expression for x

For \( \sin x + \csc x \), substituting \( \csc x = \frac{1}{\sin x} \) gives \( \sin x + \frac{1}{\sin x} \). Setting this as variable \( z \), we can rewrite it as \( z + \frac{1}{z} = 2 \) implying \( z = 1 \) since \( z + \frac{1}{z} \geq 2 \, (\text{arithmetic mean-geometric mean inequality}) \). Therefore, \( \sin x = 1 \).

Step 3: Simplify the expression for y

Next, for \( \tan y + \cot y \), we have \( \tan y = t \) and \( \cot y = \frac{1}{t} \). Substituting, we get \( t + \frac{1}{t} = 2 \), leading us to \( t = 1 \), using the same mean argument. Therefore, \( \tan y = 1 \) implying \( y = \frac{\pi}{4} \).

Step 4: Calculate \( \tan \frac{y}{2} \)

For \( y = \frac{\pi}{4} \), then \( \frac{y}{2} = \frac{\pi}{8} \). Using the half-angle identity for tangent, \( \tan \frac{y}{2} = \sqrt{\frac{1 - \cos y}{1 + \cos y}} \), and for \( y = \frac{\pi}{4} \), \( \cos y = \frac{1}{\sqrt{2}} \). Hence, \( \tan \frac{y}{2} = \sqrt{\frac{1 - \frac{1}{\sqrt{2}}}{1 + \frac{1}{\sqrt{2}}}} = \sqrt{1 - \tan^2 \frac{y}{2}} \). Solving gives \( \tan \frac{y}{2} = \frac{\sqrt{2} - 1}{\sqrt{2} + 1} \).

Step 5: Find the appropriate equation

We determined that \( \tan \frac{y}{2} = \sqrt{2} - 1 \). Simplifying this further, recognize that it must satisfy a quadratic equation. Historically manipulating, if \( \alpha = \sqrt{2} - 1 \), substitute back into provided equations. The closest match is \( \alpha^2 + 2\alpha - 1 = 0 \), given convictions from former quadrature."

Step 6: Verify the solution

Verify that substituting \( \alpha = \sqrt{2} - 1 \) into \( \alpha^2 + 2\alpha - 1 = 0 \) satisfies the equation. With calculated roots and simplifications, test ensures the equation holds true mathematically. Execute this outcome confirms the initial assertion and supports by rules of material logic.

Key Concepts

Trigonometric Identities

Trigonometric identities are fundamental tools in solving trigonometric equations. They are equations that are true for all values of the involved variables, helping simplify and solve various trigonometric expressions. Commonly used identities include the reciprocal identities like

• \( \csc x = \frac{1}{\sin x} \)
• \( \sec x = \frac{1}{\cos x} \)
• \( \cot x = \frac{1}{\tan x} \)

These identities are essential because they allow us to rewrite trigonometric functions in terms of sine, cosine, and tangent, often leading to simpler forms. For example, in the given exercise, we used the identity \( \csc x = \frac{1}{\sin x} \) to rewrite and simplify the expression \( \sin x + \csc x \). By setting it equal to 2, we deduced that \( \sin x = 1 \), simplifying the original equation. Consequently, understanding and applying these identities can significantly help solve problems involving trigonometric functions.

Half-Angle Identities

The half-angle identities are powerful in calculating the trigonometric functions of half-angles. For example, the half-angle identity for tangent is given by:

• \( \tan \frac{\theta}{2} = \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}} \)

This specific identity allows us to find the tangent of an angle not directly accessible from standard angle values. In the exercise, once it was established that \( y = \frac{\pi}{4} \), we needed to find \( \tan \frac{y}{2} \). By employing the half-angle identity, calculated due to the known value \( \cos y = \frac{1}{\sqrt{2}} \), we find that:

• \( \tan \frac{\pi}{8} = \sqrt{\frac{1 - \frac{1}{\sqrt{2}}}{1 + \frac{1}{\sqrt{2}}}} \)

Using half-angle identities simplifies evaluating trigonometric functions, particularly when dealing with angle reductions.

Quadratic Equations

Quadratic equations come in the form \( ax^2 + bx + c = 0 \). Solving these equations involves finding the values of \( x \) that make the equation true. In trigonometry, quadratic equations frequently arise when simplifying expressions or solving for specific values.
Understanding how to solve these equations using various methods, such as factoring, completing the square, or the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), is crucial. In the exercise, we established \( \tan \frac{y}{2} \) as a root of the equation \( \alpha^2 + 2\alpha - 1 = 0 \). Finding this quadratic form helped connect \( \tan \frac{y}{2} \) to a specific numerical value, needed to solve the problem with known equations.
Mastering quadratic equations enables solving more complex trigonometric problems where expressions transform into quadratic forms. Thus, recognizing and manipulating these forms is a key step in advanced trigonometric problem-solving.