Solving Trigonometry: Find tan α - cot α

This problem asks us to find the value of an expression involving tangent and cotangent of an angle α, given the value of cot α. We are given that cot α = √2 + 1, and we need to calculate tan α - cot α.

Relationship Between tan α and cot α

The tangent and cotangent of an angle are reciprocals of each other. This means that:

tan α = \frac{1}{\text{cot } \alpha}

Using this relationship, we can find the value of tan α since we know the value of cot α.

Calculating the Value of tan α

Given cot α = √2 + 1, we can find tan α:

tan \alpha = \frac{1}{\sqrt{2} + 1}

To simplify this expression, we can rationalize the denominator. We multiply the numerator and the denominator by the conjugate of the denominator, which is √2 - 1.

tan \alpha = \frac{1}{\sqrt{2} + 1} \times \frac{\sqrt{2} - 1}{\sqrt{2} - 1}

Using the difference of squares formula, (a+b)(a-b) = a^2 - b^2, the denominator becomes:

(\sqrt{2} + 1)(\sqrt{2} - 1) = (\sqrt{2})^2 - (1)^2 = 2 - 1 = 1

So, the expression for tan α simplifies to:

tan \alpha = \frac{1 \times (\sqrt{2} - 1)}{1}

tan \alpha = \sqrt{2} - 1

Calculating tan α - cot α

Now that we have the values for both tan α and cot α, we can calculate the required expression tan α - cot α.

• Value of tan α = √2 - 1
• Value of cot α = √2 + 1

Subtracting cot α from tan α:

tan \alpha - \text{cot } \alpha = (\sqrt{2} - 1) - (\sqrt{2} + 1)

Remove the parentheses, remembering to distribute the negative sign to both terms inside the second parenthesis:

tan \alpha - \text{cot } \alpha = \sqrt{2} - 1 - \sqrt{2} - 1

Group the terms with √2 and the constant terms:

tan \alpha - \text{cot } \alpha = (\sqrt{2} - \sqrt{2}) + (-1 - 1)

tan \alpha - \text{cot } \alpha = 0 + (-2)

tan \alpha - \text{cot } \alpha = -2

The value of tan α - cot α is -2.

Revision Table: Trigonometry Concepts

Let's quickly review the core trigonometric ratios involved in this problem.

Additional Information: Rationalizing Denominators

Rationalizing the denominator is a technique used to eliminate radicals from the denominator of a fraction. When the denominator is a binomial involving a square root, like a + \sqrt{b} or \sqrt{a} + \sqrt{b}, we multiply both the numerator and the denominator by its conjugate. The conjugate of a + \sqrt{b} is a - \sqrt{b}, and the conjugate of \sqrt{a} + \sqrt{b} is \sqrt{a} - \sqrt{b}. This works because the product of a binomial and its conjugate results in the difference of squares, which removes the radical (e.g., (\sqrt{x})^2 = x).

In our problem, the denominator was \sqrt{2} + 1. Its conjugate is \sqrt{2} - 1. Multiplying by the conjugate allowed us to change the denominator to a rational number (1 in this case), simplifying the expression for tan α.