Problem 198 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ x=\sec t, \quad y=\tan t $$

Step-by-Step Solution

Answer

The relationship between \(x\) and \(y\) is expressed as \(y = \sqrt{x^2 - 1}\) or \(y = -\sqrt{x^2 - 1}\).

Step 1: Recalling the Trigonometric Identities

Remember the trigonometric identity \(\sec^2 t = 1 + \tan^2 t\). This is key to finding the relation between \(x\) and \(y\).

Step 2: Substituting Given Parametric Equations

Replace \(\sec t\) and \(\tan t\) with \(x\) and \(y\) respectively in the trigonometric identity. We obtain \(x^2 = 1 + y^2\).

Step 3: Solving for y

Rearrange the equation obtained in step 2 to express \(y\) in terms of \(x\). We subtract 1 from both sides to get \(x^2 - 1 = y^2\). Then, we take the square root on both sides. Note that because \(y = \tan t\), and \(-\infty < t < \infty\), \(y\) can be either positive or negative. Therefore, we get \(y = \sqrt{x^2 - 1}\) or \(y = -\sqrt{x^2 - 1}\) depending on the value of \(t\).

Key Concepts

Trigonometric Identities

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.

Relation between x and y

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.

Solving for y

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.