cot(x) cot(y) + 1 cot(y) – cot(x) cot(x – y) =

Trigonometry (11th Edition)
11th Edition
ISBN:9780134217437
Author:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Publisher:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Chapter1: Trigonometric Functions
Section: Chapter Questions
Problem 1RE: 1. Give the measures of the complement and the supplement of an angle measuring 35°.
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### Proving the Identity in Trigonometry

We are given the following trigonometric identity to prove:

\[ \cot(x - y) = \frac{\cot(x) \cdot \cot(y) + 1}{\cot(y) - \cot(x)} \]

We'll prove this identity by using reciprocal identities and subtraction formulas step by step.

#### Step 1: Use a Reciprocal Identity, and then use a Subtraction Formula.

Starting with the left-hand side:
\[ \cot(x - y) = \frac{1}{\tan(x - y)} \]

Using the tangent subtraction formula:
\[ \tan(x - y) = \frac{\tan(x) - \tan(y)}{1 + \tan(x) \cdot \tan(y)}\]

So by reciprocal identity:
\[ \cot(x - y) = \frac{1}{\frac{\tan(x) - \tan(y)}{1 + \tan(x) \cdot \tan(y)}} \]
\[ = \frac{1 + \tan(x) \cdot \tan(y)}{\tan(x) - \tan(y)} \]

#### Step 2: Use a Reciprocal Identity, and simplify the compound fraction.

Next, express tangents in terms of cotangents:
\[ \tan(x) = \frac{1}{\cot(x)}, \quad \tan(y) = \frac{1}{\cot(y)} \]

Substituting these into the equation:
\[ \cot(x - y) = \frac{1 + \frac{1}{\cot(x)} \cdot \frac{1}{\cot(y)}}{\frac{1}{\cot(x)} - \frac{1}{\cot(y)}} \]

Simplifying the numerator and the denominator:
\[ = \frac{1 + \frac{1}{\cot(x) \cdot \cot(y)}}{\frac{1}{\cot(x)} - \frac{1}{\cot(y)}} \]
\[ = \frac{1 + \frac{1}{\cot(x) \cot(y)}}{\frac{\cot(y) - \cot(x)}{\cot(x) \cdot \cot(y)}} \]

Further simplify to:
\[ = \frac{(1 + \frac{1}{\cot(x) \cdot \cot(y)}) \cdot (\cot(x) \cdot \cot(y))}{
Transcribed Image Text:### Proving the Identity in Trigonometry We are given the following trigonometric identity to prove: \[ \cot(x - y) = \frac{\cot(x) \cdot \cot(y) + 1}{\cot(y) - \cot(x)} \] We'll prove this identity by using reciprocal identities and subtraction formulas step by step. #### Step 1: Use a Reciprocal Identity, and then use a Subtraction Formula. Starting with the left-hand side: \[ \cot(x - y) = \frac{1}{\tan(x - y)} \] Using the tangent subtraction formula: \[ \tan(x - y) = \frac{\tan(x) - \tan(y)}{1 + \tan(x) \cdot \tan(y)}\] So by reciprocal identity: \[ \cot(x - y) = \frac{1}{\frac{\tan(x) - \tan(y)}{1 + \tan(x) \cdot \tan(y)}} \] \[ = \frac{1 + \tan(x) \cdot \tan(y)}{\tan(x) - \tan(y)} \] #### Step 2: Use a Reciprocal Identity, and simplify the compound fraction. Next, express tangents in terms of cotangents: \[ \tan(x) = \frac{1}{\cot(x)}, \quad \tan(y) = \frac{1}{\cot(y)} \] Substituting these into the equation: \[ \cot(x - y) = \frac{1 + \frac{1}{\cot(x)} \cdot \frac{1}{\cot(y)}}{\frac{1}{\cot(x)} - \frac{1}{\cot(y)}} \] Simplifying the numerator and the denominator: \[ = \frac{1 + \frac{1}{\cot(x) \cdot \cot(y)}}{\frac{1}{\cot(x)} - \frac{1}{\cot(y)}} \] \[ = \frac{1 + \frac{1}{\cot(x) \cot(y)}}{\frac{\cot(y) - \cot(x)}{\cot(x) \cdot \cot(y)}} \] Further simplify to: \[ = \frac{(1 + \frac{1}{\cot(x) \cdot \cot(y)}) \cdot (\cot(x) \cdot \cot(y))}{
**Verify the Identity**

\[ \frac{1}{\sec(x) + \tan(x)} + \frac{1}{\sec(x) - \tan(x)} = 2 \sec(x) \]

To verify this trigonometric identity, we start by considering the left-hand side:

\[ \frac{1}{\sec(x) + \tan(x)} + \frac{1}{\sec(x) - \tan(x)} \]

First, find a common denominator:

\[ \frac{\sec(x) - \tan(x)}{(\sec(x) + \tan(x))(\sec(x) - \tan(x))} + \frac{\sec(x) + \tan(x)}{(\sec(x) + \tan(x))(\sec(x) - \tan(x))} \]

Combine the fractions:

\[ \frac{\sec(x) - \tan(x) + \sec(x) + \tan(x)}{(\sec(x) + \tan(x))(\sec(x) - \tan(x))} \]

Simplify the numerator:

\[ \frac{\sec(x) + \sec(x)}{\sec^2(x) - \tan^2(x)} \]

\[ \frac{2 \sec(x)}{\sec^2(x) - \tan^2(x)} \]

Noting that:

\[ \sec^2(x) - \tan^2(x) = 1 \]

The expression becomes:

\[ \frac{2 \sec(x)}{1} = 2 \sec(x) \]

Thus, we have shown that:

\[ \frac{1}{\sec(x) + \tan(x)} + \frac{1}{\sec(x) - \tan(x)} = 2 \sec(x) \]

This verifies the given identity.
Transcribed Image Text:**Verify the Identity** \[ \frac{1}{\sec(x) + \tan(x)} + \frac{1}{\sec(x) - \tan(x)} = 2 \sec(x) \] To verify this trigonometric identity, we start by considering the left-hand side: \[ \frac{1}{\sec(x) + \tan(x)} + \frac{1}{\sec(x) - \tan(x)} \] First, find a common denominator: \[ \frac{\sec(x) - \tan(x)}{(\sec(x) + \tan(x))(\sec(x) - \tan(x))} + \frac{\sec(x) + \tan(x)}{(\sec(x) + \tan(x))(\sec(x) - \tan(x))} \] Combine the fractions: \[ \frac{\sec(x) - \tan(x) + \sec(x) + \tan(x)}{(\sec(x) + \tan(x))(\sec(x) - \tan(x))} \] Simplify the numerator: \[ \frac{\sec(x) + \sec(x)}{\sec^2(x) - \tan^2(x)} \] \[ \frac{2 \sec(x)}{\sec^2(x) - \tan^2(x)} \] Noting that: \[ \sec^2(x) - \tan^2(x) = 1 \] The expression becomes: \[ \frac{2 \sec(x)}{1} = 2 \sec(x) \] Thus, we have shown that: \[ \frac{1}{\sec(x) + \tan(x)} + \frac{1}{\sec(x) - \tan(x)} = 2 \sec(x) \] This verifies the given identity.
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