Math Problem Statement

Prove that for the trigonometric equations secθ + tanθ = m and secθ - tanθ = n, the relation m² - n² = 4√(mn) holds true.

Solution

The problem presented in the image shows two equations involving trigonometric functions:

  1. secθ+tanθ=m\sec\theta + \tan\theta = m
  2. secθtanθ=n\sec\theta - \tan\theta = n

We are tasked to prove that:

m2n2=4mnm^2 - n^2 = 4\sqrt{mn}

Step 1: Use the identity for the difference of squares

We start by applying the identity m2n2=(mn)(m+n)m^2 - n^2 = (m - n)(m + n):

m2n2=(secθ+tanθ+secθtanθ)(secθ+tanθ(secθtanθ))m^2 - n^2 = (\sec\theta + \tan\theta + \sec\theta - \tan\theta)(\sec\theta + \tan\theta - (\sec\theta - \tan\theta))

This simplifies to:

m2n2=(2secθ)(2tanθ)m^2 - n^2 = (2\sec\theta)(2\tan\theta)

Thus,

m2n2=4secθtanθm^2 - n^2 = 4\sec\theta \cdot \tan\theta

Step 2: Use the identities for secant and tangent

We know from trigonometric identities that:

secθ=1cosθandtanθ=sinθcosθ\sec\theta = \frac{1}{\cos\theta} \quad \text{and} \quad \tan\theta = \frac{\sin\theta}{\cos\theta}

Therefore, secθtanθ\sec\theta \cdot \tan\theta becomes:

secθtanθ=1cosθsinθcosθ=sinθcos2θ\sec\theta \cdot \tan\theta = \frac{1}{\cos\theta} \cdot \frac{\sin\theta}{\cos\theta} = \frac{\sin\theta}{\cos^2\theta}

Thus:

m2n2=4sinθcos2θm^2 - n^2 = 4 \cdot \frac{\sin\theta}{\cos^2\theta}

Step 3: Express mn\sqrt{mn}

From the given equations m=secθ+tanθm = \sec\theta + \tan\theta and n=secθtanθn = \sec\theta - \tan\theta, the product mnmn becomes:

mn=(secθ+tanθ)(secθtanθ)=sec2θtan2θmn = (\sec\theta + \tan\theta)(\sec\theta - \tan\theta) = \sec^2\theta - \tan^2\theta

Using the Pythagorean identity sec2θtan2θ=1\sec^2\theta - \tan^2\theta = 1, we have:

mn=1mn = 1

Therefore:

mn=1=1\sqrt{mn} = \sqrt{1} = 1

Step 4: Final equation

Substituting into the equation m2n2=4mnm^2 - n^2 = 4\sqrt{mn}, we get:

m2n2=41=4m^2 - n^2 = 4\cdot 1 = 4

Thus, the equation holds true.

Let me know if you'd like more details on any of the steps, or if you have further questions.


Related questions:

  1. Can you prove the identity sec2θtan2θ=1\sec^2\theta - \tan^2\theta = 1 using basic trigonometric identities?
  2. What are the other forms of trigonometric identities for secant and tangent functions?
  3. How can we derive the formula for secθ\sec\theta and tanθ\tan\theta in terms of sine and cosine?
  4. Can you solve for θ\theta in terms of mm and nn from the given equations?
  5. How can these trigonometric equations be applied in real-world scenarios?

Tip:

When dealing with trigonometric identities, always keep basic identities like the Pythagorean identities handy. They often simplify complex expressions.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Identities
Algebraic Manipulation

Formulas

secθ + tanθ = m
secθ - tanθ = n
m² - n² = (m + n)(m - n)
sec²θ - tan²θ = 1
secθ = 1/cosθ
tanθ = sinθ/cosθ

Theorems

Pythagorean Identity

Suitable Grade Level

Grades 10-12