Proving \frac{sec^{2}A}{1-tan^{2}A} Equals sec2A

• thomas49th
In summary, the conversation is about proving the identity \frac{\sec^{2}A}{1-\tan^{2}A} = \sec2AThe person has simplified the expression to \frac{1+\frac{\sin^{2}A}{\cos^{2}A}}{1-\frac{\sin^{2}A}{\cos^{2}A}}and the book suggests it can be further simplified to \frac{\cos^{2}A+\sin^{2}A}{\cos^{2}A-\sin^{2}A}However, the person is unsure how to do this and asks for help. They are instructed to multiply both the numerator and denominator by \cos^{2
thomas49th
I have to prove this:

$$\frac{sec^{2}A}{1-tan^{2}A}\equivsec2A$$

I have got it down to this:

$$\frac{1+\frac{sin^{2}A}{cos^{2}A}}{1-\frac{sin^{2}A}{cos^{2}A}}$$

and the book says it can be then further equated to

$$\frac{cos^{2}A+sin^{2}A}{cos^{2}A-sin^{2}A}$$

but i can't see how that is done :(

Can someone show me

Thanks ;)

Multiply every term in the numerator and the denominator by cos^2 x.

What do you need to prove about it ? Generally you prove a proposition. Whats your proposition ?

There was an error in his latex code, if you know how to read latex code click on his latex and you'll see he meant

$$\frac{\sec^{2}A}{1-\tan^{2}A} = \sec2A$$

1. What is the formula for proving sec2A / (1 - tan2A) equals sec2A?

The formula for proving sec2A / (1 - tan2A) equals sec2A is based on the Pythagorean identity for secant and tangent: sec2A = 1 + tan2A. By substituting this identity into the original equation, we get (1 + tan2A) / (1 - tan2A), which simplifies to sec2A.

2. What is the Pythagorean identity for secant and tangent?

The Pythagorean identity for secant and tangent is sec2A = 1 + tan2A. This identity is derived from the Pythagorean theorem, which states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.

3. How do you prove that sec2A / (1 - tan2A) equals sec2A?

To prove that sec2A / (1 - tan2A) equals sec2A, we can use the Pythagorean identity for secant and tangent, as well as basic algebraic principles. By substituting the identity into the original equation, we can simplify it to sec2A, thus proving that the two expressions are equal.

4. Can this equation be used to solve trigonometric problems?

Yes, this equation can be used to solve trigonometric problems involving secant and tangent. It can also be used to prove other trigonometric identities and to simplify more complex expressions.

5. What are some practical applications of this equation?

This equation has practical applications in various fields, such as engineering, physics, and astronomy. It can be used to calculate angles and distances in right triangles, as well as to analyze and model periodic functions. It is also used in the study of waves and vibrations, as well as in calculating the position and movement of celestial bodies.

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