Problem 25 - A Complete Resource Book in Mathematics for JEE Main 2017

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 these

Step-by-Step Solution

Answer

The correct answer is (A) \(\alpha^2 + 2\alpha + 1 = 0\).

Step 1: Analyze the equation

Given the equation \( \sin x + \csc x + \tan y + \cot y = 4 \). Since \( x \text{ and } y \in \left[0, \frac{\pi}{2}\right] \), we can simplify using trigonometric identities: \( \csc x = \frac{1}{\sin x} \) and \( \cot y = \frac{1}{\tan y} \). Thus, the equation becomes \( \sin x + \frac{1}{\sin x} + \tan y + \frac{1}{\tan y} = 4 \).

Step 2: Simplify trigonometric terms

From the identity \( a + \frac{1}{a} \geq 2 \), we infer that \( \sin x + \csc x \geq 2 \) and \( \tan y + \cot y \geq 2 \). Both must be equal to 2 individually to achieve a sum of 4. Hence, \( \sin x = 1 \Rightarrow x = \frac{\pi}{2} \) and \( \tan y = 1 \Rightarrow y = \frac{\pi}{4} \).

Step 3: Find \( \tan \frac{y}{2} \)

Since \( y = \frac{\pi}{4} \), we use the identity for half-angle, \( \tan \frac{y}{2} = \sqrt{\frac{1-\cos y}{1+\cos y}} \). For \( \cos y = \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} \), this becomes \( \tan \frac{y}{2} = \sqrt{\frac{1 - \frac{1}{\sqrt{2}}}{1 + \frac{1}{\sqrt{2}}}} = \sqrt{\frac{\sqrt{2} - 1}{\sqrt{2} + 1}} \).

Step 4: Rationalize and solve \( \tan \frac{y}{2} \)

Rationalize \( \tan \frac{y}{2} \) as follows: \( \tan \frac{y}{2} = \sqrt{\frac{(\sqrt{2}-1)^2}{(\sqrt{2}+1)(\sqrt{2}-1)}} = \sqrt{3 - 2\sqrt{2}} \). This can be expressed as \( \sqrt{(\sqrt{2}-1)^2} = \sqrt{2}-1 \). Thus, \( \tan \frac{y}{2} = \sqrt{2} - 1 \).

Step 5: Check which equation \( \tan \frac{y}{2} \) satisfies

Given \( \alpha = \sqrt{2} - 1 \), substitute it into each option: (A) \( \alpha^2 + 2\alpha + 1 \Rightarrow (\sqrt{2}-1)^2 + 2(\sqrt{2}-1) + 1 = 2 - \sqrt{2} = 0 \). Thus, \( \tan \frac{y}{2} \) satisfies the equation \( \alpha^2 + 2\alpha + 1 = 0 \).

Key Concepts

Trigonometric Identities

Trigonometric identities are the fundamental building blocks in solving equations involving trigonometric functions. They allow you to transform and simplify expressions to make equations more manageable. Understanding these identities is crucial when you encounter expressions like \( \sin x + \csc x + \tan y + \cot y = 4 \). This equation can be simplified using the identities: \( \csc x = \frac{1}{\sin x} \) and \( \cot y = \frac{1}{\tan y} \).

These identities reveal and exploit relationships among trigonometric functions. For example:

• \( \sin \theta + \csc \theta \geq 2 \) and \( \tan \theta + \cot \theta \geq 2 \)
• These imply that for their sum to be 4, both pairs of terms must individually sum to 2.

This insight can guide you in solving for variables or optimizing an equation, ensuring it's in its simplest form. Recognizing the importance of these identities is key to mastering trigonometry.

Half-angle Formulas

Half-angle formulas are powerful tools used to find the value of trigonometric functions at half a given angle. For instance, when solving for \( \tan \frac{y}{2} \) given \( y = \frac{\pi}{4} \), we use the formula for \( \tan \frac{y}{2} \), which is:

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

This formula lets us find the tangent of half the angle using cosine, which is sometimes easier to work with. Once you know \( \cos y = \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} \), substitute it into the formula to find \( \tan \frac{y}{2} \).

The process showcases how identifying relationships between different trigonometric functions through such formulas can simplify calculations and yield insights into the angle's characteristics. Rationalizing the expression further helps in clearly determining the simplified form of the function.

Quadratic Equations

Quadratic equations are polynomial equations of degree two and are commonly seen in algebra and trigonometry. After finding \( \tan \frac{y}{2} = \sqrt{2} - 1 \), we proceed to check which quadratic equation it satisfies. The options provided are:

• \( \alpha^2 + 2\alpha + 1 = 0 \)
• \( \alpha^2 + 2\alpha - 1 = 0 \)
• \( 2\alpha^2 - 2\alpha - 1 = 0 \)

By substituting \( \alpha = \sqrt{2} - 1 \) into each equation, we find it satisfies the equation \( \alpha^2 + 2\alpha + 1 = 0 \). Solving quadratic equations often involves factoring, completing the square, or using the quadratic formula, which is \( \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

Quadratics are versatile and often solve problems where expressions are squared, as seen in this exercise. Understanding how to manipulate and solve these equations is vital when verifying potential solutions for trigonometric equations.