Problem 3 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

Solve: \(\tan ^{2} \theta+(1-\sqrt{3}) \tan \theta-\sqrt{3}=0\)

Step-by-Step Solution

Answer

\(\theta = \frac{\pi}{3} + n\pi\) or \(\theta = n\pi\), where \(n\) is an integer

Step 1: Identify Coefficients

Here, \(a=\tan^2\theta\), \(b=(1-\sqrt{3})\tan\theta\), and \(c=-\sqrt{3}\). Therefore, we have: \(a=1\), \(b=1-\sqrt{3}\), and \(c=-\sqrt{3}\).

Step 2: Apply the Quadratic Formula

Inserting the values of \(a\), \(b\), and \(c\) to the quadratic formula gives: \(\tan \theta = \frac{-(1-\sqrt{3}) \pm \sqrt{(1-\sqrt{3})^{2}-4(1)(-\sqrt{3})}}{2(1)}\). Simplify to: \(\tan \theta = \frac{-1+\sqrt{3} \pm \sqrt{-2\sqrt{3} + 4 + 3}}{2}\). This finally simplifies to: \(\tan \theta = \frac{-1+\sqrt{3} \pm \sqrt{3}}{2}\).

Step 3: Solve for \(\theta\)

This gives two possible solutions for \(\tan \theta\), depending upon the selection of either the positive or negative square root term. If we first choose the positive root, we have: \(\tan \theta = \sqrt{3}\), giving \(\theta = \frac{\pi}{3} + n\pi\), where \(n\) is an integer. If we choose the negative root, we have: \(\tan \theta = 0\), giving \(\theta = n\pi\).

Key Concepts

Quadratic Formula

The quadratic formula is an invaluable tool when solving quadratic equations, including those involving trigonometric functions. A quadratic equation typically takes the form:

• \(ax^2 + bx + c = 0\)

Here, \(a\), \(b\), and \(c\) are coefficients that determine the shape and position of the parabola represented by the equation. In trigonometric contexts, these coefficients might involve trigonometric expressions, such as \(\tan^2\theta\).

To solve a quadratic equation, we use the quadratic formula:

• \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

This formula tells us the possible values of \(x\) that satisfy the equation — essentially the "roots" or solutions of the equation. The symbol \(\pm\) indicates that there could be two different solutions: one involving the positive square root and another involving the negative square root.

In our case, \(\tan^2\theta + (1-\sqrt{3})\tan\theta - \sqrt{3} = 0\), the quadratic formula allows us to solve for \(\tan\theta\), utilizing \(a = 1\), \(b = 1-\sqrt{3}\), and \(c = -\sqrt{3}\). This applies the same fundamental rules of the quadratic formula, showing its versatility even in trigonometry.

Tangent Function

The tangent function, denoted as \(\tan\theta\), is a fundamental trigonometric function that relates the angle \(\theta\) to the ratio of the opposite side to the adjacent side in a right triangle. It can also be expressed using sine and cosine functions:

• \(\tan\theta = \frac{\sin\theta}{\cos\theta}\)

This function is periodic, meaning it repeats its values in regular intervals. The period for \(\tan\theta\) is \(\pi\), meaning that every set of solutions repeats after \(\pi\) radians.

In relation to the given equation, \(\tan\theta\) serves as an integral part of solving for \(\theta\). After applying the quadratic formula, you determine possible values for \(\tan\theta\). For instance, one solution leads to \(\tan\theta = \sqrt{3}\). Using the known reference angles, \(\tan\theta\) equals \(\sqrt{3}\) at \(\theta = \frac{\pi}{3}\), adjusting for the periodic nature of the tangent.

Because of its periodicity, we express the general solutions as:

• \(\theta = \frac{\pi}{3} + n\pi\)
• \(\theta = n\pi\)

where \(n\) is an integer, capturing all possible angles where the initial conditions hold true.

Solutions of Trigonometric Equations

Solving trigonometric equations like \(\tan^2\theta + (1-\sqrt{3})\tan\theta - \sqrt{3} = 0\) involves finding angles \(\theta\) that satisfy the equation. This would usually include methods such as algebraic manipulation, using identities, or applying quadratic solutions.

• First, identify the trigonometric form: equations can often be simplified to involve basic expressions like \(\sin\theta\), \(\cos\theta\), or \(\tan\theta\).
• Rearrange and solve using standard algebraic techniques, such as factoring or applying the quadratic formula if quadratic in nature.
• Finally, interpret the solutions in terms of the unit circle or other trigonometric identities to find specific angles \(\theta\). For the tangent function, this might involve considering angles where \(\tan\theta = 0\), as it does here.

Understanding these solutions involves recognizing their periodicity:

• For \(\tan\theta\), the solutions repeat every \(\pi\) radians.
• Thus, the solutions for \(\theta\) are often expressed in terms of \(n\pi\) or \(\frac{\pi}{3} + n\pi\), where \(n\) represents any integer.

This periodicity helps ensure all potential angles are considered, providing a complete set of solutions aligned with the characteristics of the function.