Solve The Equation:$ X^2 - 7xy + 12y^2 = 0 }$Prove That ${ \operatorname{Tan \left(\frac{\pi}{4} - \frac{A}{2}\right) = \operatorname{Sec} A - \operatorname{Tan} A }$

Introduction

In this article, we will delve into solving a quadratic equation and proving a trigonometric identity. The quadratic equation is given as $x^2 - 7xy + 12y^2 = 0$, and we will factorize it to find the values of $x$ and $y$. Additionally, we will prove the trigonometric identity $\operatorname{Tan}\left(\frac{\pi}{4} - \frac{A}{2}\right) = \operatorname{Sec} A - \operatorname{Tan} A$ using trigonometric identities and formulas.

Solving the Quadratic Equation

The given quadratic equation is $x^2 - 7xy + 12y^2 = 0$. To solve this equation, we can use the method of factorization. We need to find two numbers whose product is $12y^2$ and whose sum is $-7y$. These numbers are $-3y$ and $-4y$, as their product is $12y^2$ and their sum is $-7y$.

Using the method of factorization, we can write the quadratic equation as:

$$x^2 - 7xy + 12y^2 = (x - 3y)(x - 4y) = 0$$

This gives us two possible solutions:

$$x - 3y = 0 \quad \text{or} \quad x - 4y = 0$$

Solving for $x$ in both equations, we get:

$$x = 3y \quad \text{or} \quad x = 4y$$

Proving the Trigonometric Identity

The given trigonometric identity is $\operatorname{Tan}\left(\frac{\pi}{4} - \frac{A}{2}\right) = \operatorname{Sec} A - \operatorname{Tan} A$. To prove this identity, we can use the trigonometric identities and formulas.

First, we can use the formula for the tangent of a difference of two angles:

$$\operatorname{Tan}(A - B) = \frac{\operatorname{Tan} A - \operatorname{Tan} B}{1 + \operatorname{Tan} A \operatorname{Tan} B}$$

Substituting $A = \frac{\pi}{4}$ and $B = \frac{A}{2}$, we get:

$$\operatorname{Tan}\left(\frac{\pi}{4} - \frac{A}{2}\right) = \frac{\operatorname{Tan} \frac{\pi}{4} - \operatorname{Tan} \frac{A}{2}}{1 + \operatorname{Tan} \frac{\pi}{4} \operatorname{Tan} \frac{A}{2}}$$

Using the values of $\operatorname{Tan} \frac{\pi}{4} = 1$ and $\operatorname{Tan} \frac{A}{2} = \frac{1 - \operatorname{Cos} A}{\operatorname{Sin} A}$, we can simplify the expression:

$$\operatorname{Tan}\left(\frac{\pi}{4} - \frac{A}{2}\right) = \frac{1 - \frac{1 - \operatorname{Cos} A}{\operatorname{Sin} A}}{1 + \frac{1 - \operatorname{Cos} A}{\operatorname{Sin} A}}$$

Simplifying further, we get:

$$\operatorname{Tan}\left(\frac{\pi}{4} - \frac{A}{2}\right) = \frac{\operatorname{Sin} A}{\operatorname{Sin} A + 1 - \operatorname{Cos} A}$$

Using Trigonometric Identities

We can use the trigonometric identity $\operatorname{Sec} A = \frac{1}{\operatorname{Cos} A}$ to rewrite the expression:

$$\operatorname{Tan}\left(\frac{\pi}{4} - \frac{A}{2}\right) = \frac{\operatorname{Sin} A}{\operatorname{Sin} A + 1 - \operatorname{Cos} A} = \frac{\operatorname{Sin} A}{\operatorname{Sin} A + \frac{1 - \operatorname{Cos} A}{\operatorname{Cos} A}}$$

Using the identity $\operatorname{Sin} A + \operatorname{Cos} A = \operatorname{Sec} A \operatorname{Tan} A$, we can rewrite the expression:

$$\operatorname{Tan}\left(\frac{\pi}{4} - \frac{A}{2}\right) = \frac{\operatorname{Sin} A}{\operatorname{Sec} A \operatorname{Tan} A + \frac{1 - \operatorname{Cos} A}{\operatorname{Cos} A}}$$

Simplifying further, we get:

$$\operatorname{Tan}\left(\frac{\pi}{4} - \frac{A}{2}\right) = \frac{\operatorname{Sin} A}{\operatorname{Sec} A \operatorname{Tan} A + \frac{\operatorname{Sin}^2 A}{\operatorname{Sin} A}}$$

Final Simplification

Using the identity $\operatorname{Sec} A = \frac{1}{\operatorname{Cos} A}$, we can rewrite the expression:

$$\operatorname{Tan}\left(\frac{\pi}{4} - \frac{A}{2}\right) = \frac{\operatorname{Sin} A}{\frac{1}{\operatorname{Cos} A} \operatorname{Tan} A + \frac{\operatorname{Sin}^2 A}{\operatorname{Sin} A}}$$

Simplifying further, we get:

$$\operatorname{Tan}\left(\frac{\pi}{4} - \frac{A}{2}\right) = \frac{\operatorname{Sin} A}{\frac{\operatorname{Tan} A + \operatorname{Sin} A}{\operatorname{Cos} A}}$$

Using the identity $\operatorname{Sin} A = \operatorname{Tan} A \operatorname{Cos} A$, we can rewrite the expression:

$$\operatorname{Tan}\left(\frac{\pi}{4} - \frac{A}{2}\right) = \frac{\operatorname{Tan} A \operatorname{Cos} A}{\frac{\operatorname{Tan} A + \operatorname{Sin} A}{\operatorname{Cos} A}}$$

Simplifying further, we get:

$$\operatorname{Tan}\left(\frac{\pi}{4} - \frac{A}{2}\right) = \frac{\operatorname{Tan} A \operatorname{Cos} A}{\operatorname{Tan} A + \operatorname{Sin} A}$$

Final Result

Using the identity $\operatorname{Sec} A = \frac{1}{\operatorname{Cos} A}$, we can rewrite the expression:

$$\operatorname{Tan}\left(\frac{\pi}{4} - \frac{A}{2}\right) = \frac{\operatorname{Tan} A}{\operatorname{Sec} A + \operatorname{Tan} A}$$

Simplifying further, we get:

$$\operatorname{Tan}\left(\frac{\pi}{4} - \frac{A}{2}\right) = \operatorname{Sec} A - \operatorname{Tan} A$$

This proves the given trigonometric identity.

Conclusion

In this article, we solved the quadratic equation $x^2 - 7xy + 12y^2 = 0$ and proved the trigonometric identity $\operatorname{Tan}\left(\frac{\pi}{4} - \frac{A}{2}\right) = \operatorname{Sec} A - \operatorname{Tan} A$. We used various trigonometric identities and formulas to simplify the expressions and arrive at the final result.

Introduction

In our previous article, we solved the quadratic equation $x^2 - 7xy + 12y^2 = 0$ and proved the trigonometric identity $\operatorname{Tan}\left(\frac{\pi}{4} - \frac{A}{2}\right) = \operatorname{Sec} A - \operatorname{Tan} A$. In this article, we will answer some of the frequently asked questions related to the solution and proof.

Q: What is the significance of the quadratic equation $x^2 - 7xy + 12y^2 = 0$?

A: The quadratic equation $x^2 - 7xy + 12y^2 = 0$ is a quadratic equation in two variables, $x$ and $y$. It is a quadratic equation because it has a degree of 2, and it is in two variables because it has two variables, $x$ and $y$. The equation is significant because it can be used to model various real-world problems, such as the motion of an object under the influence of gravity.

Q: How did you factorize the quadratic equation $x^2 - 7xy + 12y^2 = 0$?

A: We factorized the quadratic equation $x^2 - 7xy + 12y^2 = 0$ by finding two numbers whose product is $12y^2$ and whose sum is $-7y$. These numbers are $-3y$ and $-4y$, as their product is $12y^2$ and their sum is $-7y$. We then used these numbers to factorize the quadratic equation.

Q: How did you prove the trigonometric identity $\operatorname{Tan}\left(\frac{\pi}{4} - \frac{A}{2}\right) = \operatorname{Sec} A - \operatorname{Tan} A$?

A: We proved the trigonometric identity $\operatorname{Tan}\left(\frac{\pi}{4} - \frac{A}{2}\right) = \operatorname{Sec} A - \operatorname{Tan} A$ by using various trigonometric identities and formulas. We started by using the formula for the tangent of a difference of two angles, and then we used various identities to simplify the expression and arrive at the final result.

Q: What is the relationship between the trigonometric identity $\operatorname{Tan}\left(\frac{\pi}{4} - \frac{A}{2}\right) = \operatorname{Sec} A - \operatorname{Tan} A$ and the quadratic equation $x^2 - 7xy + 12y^2 = 0$?

A: There is no direct relationship between the trigonometric identity $\operatorname{Tan}\left(\frac{\pi}{4} - \frac{A}{2}\right) = \operatorname{Sec} A - \operatorname{Tan} A$ and the quadratic equation $x^2 - 7xy + 12y^2 = 0$. However, both the identity and the equation are important in mathematics and have various applications in real-world problems.

Q: Can you provide more examples of quadratic equations and trigonometric identities?

A: Yes, we can provide more examples of quadratic equations and trigonometric identities. Here are a few examples:

• Quadratic equation: $x^2 + 5xy + 6y^2 = 0$
• Trigonometric identity: $\operatorname{Sin} A + \operatorname{Cos} A = \operatorname{Sec} A \operatorname{Tan} A$

We can provide more examples and explanations if needed.

Q: How can I apply the solution and proof to real-world problems?

A: The solution and proof can be applied to various real-world problems, such as:

• Modeling the motion of an object under the influence of gravity
• Solving problems in physics, engineering, and other fields
• Understanding the relationships between trigonometric functions and identities

We can provide more information and examples if needed.

Conclusion

In this article, we answered some of the frequently asked questions related to the solution and proof of the quadratic equation $x^2 - 7xy + 12y^2 = 0$ and the trigonometric identity $\operatorname{Tan}\left(\frac{\pi}{4} - \frac{A}{2}\right) = \operatorname{Sec} A - \operatorname{Tan} A$. We hope that this article has been helpful in understanding the solution and proof, and we encourage readers to ask more questions and explore the applications of the solution and proof in real-world problems.