Question

If \(x=\sec \theta+\tan \theta, y=\sec \theta-\tan \theta\), then the relation between \(x\) and \(y\) is : (a) \(x^{2}+y^{2}=0\) (b) \(x^{2}=y^{2}\) (c) \(x^{2}=y\) (d) \(x y=1\)

Short answer

Answer: The correct relationship between x and y is given by \(xy = 1\).

Step-by-step solution

Identifying the given equations

Given \(x = \sec{θ} + \tan{θ}\) and \(y = \sec{θ} - \tan{θ}\), we will use these equations to eliminate θ.

Isolate θ using one of the equations

Let's isolate θ using the first equation, \(x = \sec{θ} + \tan{θ}\). We can write \(\tan{θ}\) in terms of \(\sec{θ}\) using the identity \(\tan{θ} = \sqrt{\sec^2{θ} - 1}\) (derived from the Pythagorean identity). The equation becomes:\(x = \sec{θ} + \sqrt{\sec^2{θ} - 1}\).

Square the equation

To eliminate the square root, we are going to square the equation above:\(x^2 = (\sec{θ} + \sqrt{\sec^2{θ} - 1})^2\).

Simplify the squared equation

Simplify and expand the squared equation:\(x^2 = \sec^2{θ} + 2\sec{θ}\sqrt{\sec^2{θ} - 1} + (\sec^2{θ} - 1)\).

Isolate the square root term

We want to isolate the square root term, so let's do that:\(2\sec{θ}\sqrt{\sec^2{θ} - 1} = x^2 - 2\sec^2{θ} + 1\).

Square the equation again

Now, let's square the equation again to eliminate the remaining square root term:\((2\sec{θ}\sqrt{\sec^2{θ} - 1})^2 = (x^2 - 2\sec^2{θ} + 1)^2\).

Simplify and cancel out terms

Simplify and cancel out terms on both sides of the equation to get:\(4\sec^2{θ}(\sec^2{θ} - 1) = (x^2 - 2\sec^2{θ} + 1)^2\).

Express the equation in terms of x and y

Next, let's express the equation in terms of x and y. Recall that \(x = \sec{θ} + \tan{θ}\) and \(y = \sec{θ} - \tan{θ}\). We have:\(4(\sec^2{θ} - 1)(\sec^2{θ} - \tan^2{θ}) = 4(xy)^2\).

Simplify the equation

Simplify the equation to get:\(4\sec^2{θ}\tan^2{θ} - 4\tan^2{θ} = 4(xy)^2\). Divide both sides by 4:\(\sec^2{θ}\tan^2{θ} - \tan^2{θ} = (xy)^2\).

Determine the relationship between x and y

From the above equation, we can deduce that \(xy = 1\). So, the correct relationship between x and y is given by answer choice (d): \(xy = 1\).

Key concepts

Secant and Tangent Relationships

Understanding the relationship between secant and tangent functions is essential in trigonometry. The secant function, denoted as \(\sec{\theta}\), is defined as the reciprocal of the cosine function, whereas the tangent function, \(\tan{\theta}\), is the quotient of the sine and cosine functions. A powerful relationship between these two functions is derived from the Pythagorean identity, stating \(\tan^2{\theta} + 1 = \sec^2{\theta}\). Rewrite this, and you get \(\tan{\theta} = \sqrt{\sec^2{\theta} - 1}\), assuming \(\theta\) is in the correct quadrant for square roots to be valid.

When manipulating equations involving secant and tangent, it often involves squaring both sides to eliminate square roots. As shown in the given exercise, by strategically expressing tangent in terms of secant and vice versa, it is possible to simplify complex expressions and ascertain relationships between different trigonometric expressions. This technique, apart from being a fundamental trigonometric skill, is crucial in solving various mathematical problems, including algebraic and calculus problems that involve trigonometric functions.

Pythagorean Identity

The Pythagorean identity is one of the most pivotal concepts in trigonometry. It is based on the Pythagorean theorem's link to the unit circle. In its most basic form, the identity states \(\sin^2{\theta} + \cos^2{\theta} = 1\). Since \(\sec{\theta}\) is the reciprocal of \(\cos{\theta}\), the identity can be restated as \(\tan^2{\theta} + 1 = \sec^2{\theta}\).

This identity is not just a formula to memorize; it is a fundamental tool that helps simplify and solve many trigonometric equations. For instance, in our exercise, the Pythagorean identity is employed to establish a relationship between secant and tangent, which then facilitates the elimination of the variable \(\theta\), allowing us to relate \(x\) and \(y\) directly without needing to solve for \(\theta\) individually.

Solving Trigonometric Equations

Trigonometric equations may seem daunting, but with the right strategies, they can be approached systematically. Solving these equations often hinges on employing trigonometric identities to simplify the expressions and isolate variables like done in our exercise. In practice, squaring both sides of an equation is a common step, but one must be cautious, as this can introduce extraneous solutions that do not satisfy the original equation.

Once an equation is simplified, you might need to perform operations such as factoring, expanding, or employing other algebraic techniques to solve for the unknowns. In the context of our exercise, the clever use of identities and algebra helps us bypass directly solving for \(\theta\) and instead directly find a relationship between \(x\) and \(y\). Remember that verifying your solutions against the original equations is an excellent practice to ensure no extraneous solutions have been introduced during the process.