Math Problem Statement

1/cot theta(sec theta -1) + tan theta /(sec theta +1), is equal to

Solution

To simplify the given expression, we start with the original expression:

1cotθ(secθ1)+tanθsecθ+1\frac{1}{\cot \theta (\sec \theta - 1)} + \frac{\tan \theta}{\sec \theta + 1}

Step 1: Express in terms of sine and cosine

Recall that: cotθ=cosθsinθ,tanθ=sinθcosθ,andsecθ=1cosθ\cot \theta = \frac{\cos \theta}{\sin \theta}, \quad \tan \theta = \frac{\sin \theta}{\cos \theta}, \quad \text{and} \quad \sec \theta = \frac{1}{\cos \theta}

Substitute these into the expression:

1cosθsinθ(1cosθ1)+sinθcosθ1cosθ+1\frac{1}{\frac{\cos \theta}{\sin \theta} \left(\frac{1}{\cos \theta} - 1\right)} + \frac{\frac{\sin \theta}{\cos \theta}}{\frac{1}{\cos \theta} + 1}

Simplify the terms inside the fractions:

  1. For the first term:

1cosθsinθ(1cosθcosθ)=1cosθ(1cosθ)sinθcosθ=sinθ1cosθ\frac{1}{\frac{\cos \theta}{\sin \theta} \left(\frac{1 - \cos \theta}{\cos \theta}\right)} = \frac{1}{\frac{\cos \theta (1 - \cos \theta)}{\sin \theta \cos \theta}} = \frac{\sin \theta}{1 - \cos \theta}

  1. For the second term:

sinθcosθ1+cosθcosθ=sinθ1+cosθ\frac{\frac{\sin \theta}{\cos \theta}}{\frac{1 + \cos \theta}{\cos \theta}} = \frac{\sin \theta}{1 + \cos \theta}

Now the expression becomes:

sinθ1cosθ+sinθ1+cosθ\frac{\sin \theta}{1 - \cos \theta} + \frac{\sin \theta}{1 + \cos \theta}

Step 2: Combine the fractions

To add the two fractions, we need a common denominator:

sinθ(1+cosθ)+sinθ(1cosθ)(1cosθ)(1+cosθ)\frac{\sin \theta (1 + \cos \theta) + \sin \theta (1 - \cos \theta)}{(1 - \cos \theta)(1 + \cos \theta)}

Simplify the numerator:

sinθ(1+cosθ)+sinθ(1cosθ)=sinθ(1+cosθ+1cosθ)=sinθ(2)\sin \theta (1 + \cos \theta) + \sin \theta (1 - \cos \theta) = \sin \theta (1 + \cos \theta + 1 - \cos \theta) = \sin \theta (2)

So, the expression simplifies to:

2sinθ1cos2θ\frac{2\sin \theta}{1 - \cos^2 \theta}

Since 1cos2θ=sin2θ1 - \cos^2 \theta = \sin^2 \theta (using the Pythagorean identity), we get:

2sinθsin2θ=2sinθ\frac{2\sin \theta}{\sin^2 \theta} = \frac{2}{\sin \theta}

Final Answer:

2sinθ=2cscθ\frac{2}{\sin \theta} = 2 \csc \theta

So, the given expression simplifies to 2cscθ\boxed{2 \csc \theta}.

Would you like further details or have any questions?

Here are 5 related questions:

  1. How do you simplify trigonometric identities using Pythagorean identities?
  2. What are the reciprocal trigonometric functions, and how are they used in simplifications?
  3. Can you explain the process of finding a common denominator in rational expressions?
  4. What are some key trigonometric identities that are frequently used in simplifications?
  5. How can we express other trigonometric functions in terms of sine and cosine?

Tip: Always check for Pythagorean identities when simplifying trigonometric expressions, as they often provide a simpler form.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Trigonometric Identities

Formulas

Pythagorean identities

Theorems

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Suitable Grade Level

Grades 11-12