Problem 28 - A Complete Resource Book in Mathematics for JEE Main

Question

A solution of the equation \((1-\tan \theta)(1+\tan \theta) \sec ^{2} \theta\) \(+2 \tan ^{2} \theta=0\), where \(\theta\) lies in the interval \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\) is given by (A) \(\theta=0\) (B) \(\theta=\frac{\pi}{3}\) or \(-\frac{\pi}{3}\) (C) \(\theta=\frac{\pi}{6}\) (D) \(\theta=-\frac{\pi}{6}\)

Step-by-Step Solution

Answer

The solutions are \(\theta=\frac{\pi}{3}\) and \(\theta=-\frac{\pi}{3}\). Option (B) is correct.

Step 1: Simplify the Expression

Start by rewriting the given equation \((1-\tan \theta)(1+\tan \theta) \sec^{2} \theta + 2 \tan^{2} \theta = 0\).The expression \((1-\tan \theta)(1+\tan \theta)\) equals \(1-\tan^{2} \theta\) using the difference of squares. Substitute this into the equation to get \( (1-\tan^{2} \theta) \sec^{2} \theta + 2 \tan^{2} \theta = 0\).

Step 2: Substitute Trigonometric Identity

Recall the identity \(\sec^2 \theta = 1 + \tan^2 \theta\). Substitute \(\sec^2 \theta\) into the equation:\((1-\tan^2 \theta)(1+\tan^2 \theta) + 2\tan^2 \theta = 0\).This simplifies to \( 1 - \tan^4 \theta + 2\tan^2 \theta = 0\).

Step 3: Form a Quadratic Equation

Let \(x = \tan^2 \theta\), then the equation becomes \(1 - x^2 + 2x = 0\).Rearrange the terms to form a quadratic equation:\(x^2 - 2x - 1 = 0\).

Step 4: Solve the Quadratic Equation

Solve the quadratic equation \(x^2 - 2x - 1 = 0\) using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \(a = 1, b = -2, c = -1\). Substitute these values to get:\[ x = \frac{2 \pm \sqrt{4 + 4}}{2} \]\[ x = \frac{2 \pm \sqrt{8}}{2} \]\[ x = \frac{2 \pm 2\sqrt{2}}{2} \]\[ x = 1 \pm \sqrt{2} \].Since \(x = \tan^2 \theta\) cannot be negative, we only consider \(x = 1 + \sqrt{2}\).

Step 5: Calculate Possible Values for θ

Since \(x = \tan^2 \theta = 1 + \sqrt{2}\),\(\tan \theta = \pm \sqrt{1 + \sqrt{2}}\).Using this value, calculate possible angles for \(\theta\):- Check \(\tan(\frac{\pi}{3}) = \sqrt{3}\),- Check \(\tan(-\frac{\pi}{3}) = -\sqrt{3}\).These values match since\(\sqrt{3} \approx \sqrt{1 + \sqrt{2}}\).

Step 6: Verify Interval Validity

The problem states that \(\theta\) is within the interval \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\).The angles \(\theta = \frac{\pi}{3}\) and \(\theta = -\frac{\pi}{3}\) fall within this interval, thus confirming their validity in our solution.

Key Concepts

Trigonometric Identities

In trigonometry, identities are equations that hold true for all values of the involved variables. Understanding these identities is crucial while solving trigonometric equations. One important identity used in this problem is the difference of squares, which states that \((a-b)(a+b) = a^2-b^2\). For instance, the expression \((1 - \tan \theta)(1 + \tan \theta)\) simplifies to \(1 - \tan^2 \theta\).

Another key identity is \(\sec^2 \theta = 1 + \tan^2 \theta\), which relates the terms involving tangent and secant. This particular identity is derived from the Pythagorean identity \(\sin^2 \theta + \cos^2 \theta = 1\), noting that \(\sec \theta = \frac{1}{\cos \theta}\). It transforms complex expressions into simpler, more manageable forms, easing the overall solving process. Understanding these identities allows us to simplify the given trigonometric equation effectively.

Quadratic Equations

Quadratic equations are polynomial equations of degree two, typically expressed as \(ax^2 + bx + c = 0\). Solving these equations is a fundamental aspect of algebra and can often be applied in solving trigonometric problems, as demonstrated here. In the given problem, we simplify the equation to a form \(x^2 - 2x - 1 = 0\) by letting \(x = \tan^2 \theta\).

To solve this quadratic equation, we can use the quadratic formula:

• Identify coefficients: \(a = 1\), \(b = -2\), \(c = -1\).
• Apply the formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
• Substitute values and simplify.

Using the quadratic formula, we find \(x = 1 \pm \sqrt{2}\). However, since \(x = \tan^2 \theta\) must be non-negative, we only consider \(x = 1 + \sqrt{2}\). This value of \(x\) leads us to possible values for \(\tan \theta\) and, consequently, for \(\theta\).

Interval Validity

Interval validity is an important consideration when solving trigonometric equations because it ensures that the solutions satisfy the domain conditions of the problem. In this problem, we need to find the values of \(\theta\) that lie within the interval \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\).

To check the validity of solutions within a specified interval,

• Calculate the potential angles for \(\theta\) using \(\tan \theta = \pm \sqrt{1 + \sqrt{2}}\).
• Verify whether these calculated angles fit within the given interval.

The angles \(\theta = \frac{\pi}{3}\) and \(\theta = -\frac{\pi}{3}\) are checked, and they indeed fall within the specified interval. This verification step is essential to ensure that the solutions not only satisfy the trigonometric equation but also adhere to the domain constraints provided in the problem.