### Proving the Identity in TrigonometryWe 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.

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Description: ### Proving the Identity in TrigonometryWe 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 formul

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Chapter1: Trigonometric Functions

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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))}{

**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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