Question

If \(\tan \theta_{1}, \tan \theta_{2}\) be the roots of \(a \tan \theta+b \sec \theta=c\), then \(\tan ^{-1} \frac{2 a c}{\left(a^{2}-c^{2}\right)}\) is equal to (a) \(2\left(\theta_{1}-\theta_{2}\right)\) (b) \(2\left(\theta_{1}+\theta_{2}\right)\) (c) \(\theta_{1}+\theta_{2}\) (d) \(\theta_{1}-\theta_{2}\)

Short answer

Question: Express the sum or difference of the angles of the roots of the equation \(a\tan\theta + b\sec\theta = c\) in terms of the expression \(\tan^{-1} \frac{2ac}{a^2 - c^2}\). Answer: The sum of the angles of the roots of the equation can be expressed as \(\theta_1 + \theta_2 = \tan^{-1} \frac{2ac}{a^2 - c^2}\).

Step-by-step solution

Write down the given equation and expression

We are given the equation: \[a \tan\theta + b \sec\theta = c \] and we need to find the relationship between the given expression: \[\tan^{-1} \frac{2ac}{a^2 - c^2} \]

Express secant in terms of tangent

Rewrite the \(\sec\theta\) in the given equation using the identity \(\sec^2\theta = 1 + \tan^2\theta\): \[\sec\theta = \sqrt{1 + \tan^2\theta} \] Now substitute this into the given equation: \[a \tan\theta + b \sqrt{1 + \tan^2\theta} = c\]

Make a substitution

Let \(x = \tan\theta\). Then the equation becomes: \[a x + b \sqrt{1 + x^2} = c\]

Solve for x

To solve for \(x\), first isolate \(\sqrt{1 + x^2}\): \[b \sqrt{1 + x^2} = c - ax\] Now, square both sides of the equation: \[b^2(1+x^2) = (c-ax)^2\]

Factorize and simplify

Expand and rearrange the equation: \[a^2x^2 + 2acx - (a^2 - b^2) = 0\] Now, let's notice that the expression we want to relate is: \[\tan^{-1} \frac{2ac}{a^2 - c^2}\]

Relate the given expression to the quadratic equation

The given expression is related to the sum of the roots of the quadratic equation obtained in Step 5. Since \(x = \tan\theta\), the sum of the roots of the quadratic equation represents the sum of the tangent functions of the angles. Using the sum of roots formula for a quadratic equation, we can find the relationship: \[x_1 + x_2 = \tan\theta_1 + \tan\theta_2 = - \frac{2ac}{a^2}\]

Determine the relationship with the tangent inverse function

We have: \[\tan^{-1}(\tan\theta_1 + \tan\theta_2) = \tan^{-1} \frac{2ac}{a^2 - c^2}\] Since the sum of the roots is equal to the expression inside the tangent inverse function, we have the relationship: \[\theta_1 + \theta_2 = \tan^{-1} \frac{2ac}{a^2 - c^2}\] Thus, the answer is: (c) \(\theta_1 + \theta_2\)

Key concepts

Tangents and Secants

Understanding the relationship between tangents and secants is essential in solving trigonometric equations. Tangent, denoted as \( \tan \theta \), is a basic trigonometric function defined as the ratio of the opposite side to the adjacent side in a right triangle. Secant, or \( \sec \theta \), is another trigonometric function representing the reciprocal of the cosine function, given by \( \sec \theta = \frac{1}{\cos \theta} \). This means that:

• \( \tan \theta = \frac{\sin \theta}{\cos \theta} \)
• \( \sec \theta = \frac{1}{\cos \theta} \)

In the given exercise, we explore the connection between these functions by expressing \( \sec \theta \) in terms of \( \tan \theta \) through the identity \( \sec^2 \theta = 1 + \tan^2 \theta \). This identity is derived from the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \), which helps us transform complex trigonometric equations into solvable forms. By substituting \( \sec \theta = \sqrt{1 + \tan^2 \theta} \) into the equation, we can then manipulate it to solve for \( \tan \theta \). This relationship between tangent and secant provides a foundation for delving into more complex problem-solving strategies.

Quadratic Equations

Quadratic equations are fundamental in algebra and appear frequently in trigonometry as well. A quadratic equation is generally of the form \( ax^2 + bx + c = 0 \). In the exercise, our task involves transforming a trigonometric equation into a quadratic equation by using \( x = \tan \theta \). This substitution aids in simplifying the expression and results in an equation that resembles the standard quadratic form:

• \( a^2 x^2 + 2acx - (a^2 - b^2) = 0 \)

To solve such an equation, we often use the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), which provides the roots of the equation. Here, \( \tan \theta_1 \) and \( \tan \theta_2 \) are roots. Understanding the sum and product of the roots is crucial as it helps establish relationships between the unknowns and trigonometric expressions. For the sum of roots \( (\theta_1 + \theta_2) \), we use the relation from the quadratic equation properties: \( \theta_1 + \theta_2 = -\frac{b}{a} \). This facilitates the connection between algebraic solutions and trigonometric expressions.

Inverse Trigonometric Functions

Inverse trigonometric functions provide angles given specific trigonometric function values. They are denoted as \( \tan^{-1}, \sin^{-1}, \) and \( \cos^{-1} \), respectively, for tangent, sine, and cosine functions. These functions are critical in translating between angle measures and their trigonometric functions in problems. In the exercise, we've used the inverse tangent function, \( \tan^{-1} \), to determine the angle equivalent of a given trigonometric ratio.

• The expression \( \tan^{-1} \frac{2ac}{a^2 - c^2} \) seeks to recover the angle from the ratio derived from the solution.
• This step allows us to map the algebraic expression to its trigonometric origins, integrating our understanding of trigonometric identities and equations.

Recognizing the role of inverse functions is vital in trigonometry problems since it allows us to step back to the angle measure, offering a bridge between numeric values and their trigonometric representations. This process helps confirm if our solutions correspond to meaningful angles, such as \( \theta_1 + \theta_2 \).