91Ó°ÊÓ

Trigonometric identities are equations involving trigonometric functions that hold true for all values within their domains. One important identity is \( \sec^2 t = 1 + \tan^2 t \). This identity is crucial as it connects the secant and tangent functions, allowing us to link different trigonometric expressions.

This particular identity is derived from the Pythagorean identity \( \sin^2 t + \cos^2 t = 1 \). By dividing throughout this equation by \( \cos^2 t \), we obtain another form: \( \tan^2 t + 1 = \sec^2 t \). This simplifies many calculations, making it easier to transition between trigonometric functions in equations.

Understanding these identities helps in simplifying complex equations involving trigonometric terms, providing a more straightforward way to solve them.

In parametric equations, variables such as \( x \) and \( y \) are expressed in terms of a third variable, often \( t \), which is called the parameter. Here, we have the parametric equations: \( x = \sec t \) and \( y = \tan t \). Our goal is to find a direct relationship between \( x \) and \( y \) without the parameter \( t \).

By using the identity \( \sec^2 t = 1 + \tan^2 t \), we substitute \( x \) and \( y \) for \( \sec t \) and \( \tan t \) respectively. This gives us the equation \( x^2 = 1 + y^2 \).

This equation is the desired relation between \( x \) and \( y \). It allows us to understand how changes in one of these variables affect the other, eliminating the parameter \( t \) and simplifying the problem into two variables.

To solve for \( y \) in the relation \( x^2 = 1 + y^2 \), we need to express \( y \) solely in terms of \( x \). Start by subtracting \( 1 \) from both sides to isolate the \( y^2 \) term, resulting in \( y^2 = x^2 - 1 \).

Next, take the square root of both sides. This gives us \( y = \sqrt{x^2 - 1} \) and \( y = -\sqrt{x^2 - 1} \). The two solutions arise because squaring either a positive or negative number results in the same number, thus both square roots are valid.

It's essential to remember that \( y = \tan t \), which means that \( y \) can take on any real value within its domain. Understanding this solution provides the flexibility to choose the correct value of \( y \) that fits the context of the problem, be it the positive or negative root.