Understanding the Trigonometric Relationship

The question asks us to find the value of the expression \(\tan \alpha - \cot \alpha\), given that \(\tan \alpha = \sqrt{2} - 1\). To solve this, we need to use the fundamental relationship between the tangent and cotangent trigonometric functions.

The cotangent of an angle is the reciprocal of the tangent of the same angle. This means:

\[ \cot \alpha = \frac{1}{\tan \alpha} \]

Given the value of \(\tan \alpha\), we can easily find the value of \(\cot \alpha\).

Calculating cotα from tanα

We are given \(\tan \alpha = \sqrt{2} - 1\). Using the reciprocal relationship:

\[ \cot \alpha = \frac{1}{\sqrt{2} - 1} \]

To simplify this expression and make it easier to work with, we should rationalize the denominator. We do this by multiplying both the numerator and the denominator by the conjugate of the denominator, which is \(\sqrt{2} + 1\). The conjugate is formed by changing the sign between the terms.

\[ \cot \alpha = \frac{1}{\sqrt{2} - 1} \times \frac{\sqrt{2} + 1}{\sqrt{2} + 1} \]

Now, we multiply the numerators and the denominators:

Numerator: \(1 \times (\sqrt{2} + 1) = \sqrt{2} + 1\)

Denominator: \((\sqrt{2} - 1)(\sqrt{2} + 1)\)

The denominator is in the form of \((a-b)(a+b)\), which simplifies to \(a^2 - b^2\). Here, \(a = \sqrt{2}\) and \(b = 1\).

So, the denominator is: \((\sqrt{2})^2 - (1)^2 = 2 - 1 = 1\)

Putting the numerator and denominator back together:

\[ \cot \alpha = \frac{\sqrt{2} + 1}{1} = \sqrt{2} + 1 \]

So, we have found that \(\cot \alpha = \sqrt{2} + 1\).

Evaluating tanα – cotα

Now that we have the values for both \(\tan \alpha\) and \(\cot \alpha\), we can substitute them into the expression \(\tan \alpha - \cot \alpha\).

We were given \(\tan \alpha = \sqrt{2} - 1\).

We calculated \(\cot \alpha = \sqrt{2} + 1\).

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

Carefully remove the parentheses. Remember to distribute the negative sign to both terms inside the second set of parentheses:

\[ \tan \alpha - \cot \alpha = \sqrt{2} - 1 - \sqrt{2} - 1 \]

Now, group the like terms (the terms with \(\sqrt{2}\) and the constant terms):

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

Simplify each group:

• \(\sqrt{2} - \sqrt{2} = 0\)
• \(-1 - 1 = -2\)

So, the expression simplifies to:

\[ \tan \alpha - \cot \alpha = 0 + (-2) = -2 \]

The value of \(\tan \alpha - \cot \alpha\) is -2.

Summary of Steps

Here's a quick recap of the steps taken to solve the problem:

1. Identify the given value of \(\tan \alpha\).
2. Use the reciprocal identity \(\cot \alpha = \frac{1}{\tan \alpha}\) to find \(\cot \alpha\).
3. Rationalize the denominator when calculating \(\cot \alpha\).
4. Substitute the values of \(\tan \alpha\) and \(\cot \alpha\) into the required expression \(\tan \alpha - \cot \alpha\).
5. Simplify the expression by combining like terms.

Revision Table: Trigonometry Basics

Let's quickly review some basic trigonometric identities that are useful for solving such problems.

Additional Information: Rationalizing Denominators

Rationalizing the denominator is a technique used to eliminate radicals (like square roots) from the denominator of a fraction. This makes expressions easier to work with and is often considered standard form.

When the denominator is a binomial involving a square root, like \(a - \sqrt{b}\) or \(\sqrt{a} - \sqrt{b}\), you multiply both the numerator and the denominator by its conjugate. The conjugate is formed by changing the sign in the middle.

• The conjugate of \(a + \sqrt{b}\) is \(a - \sqrt{b}\).
• The conjugate of \(a - \sqrt{b}\) is \(a + \sqrt{b}\).
• The conjugate of \(\sqrt{a} + \sqrt{b}\) is \(\sqrt{a} - \sqrt{b}\).
• The conjugate of \(\sqrt{a} - \sqrt{b}\) is \(\sqrt{a} + \sqrt{b}\).

The product of a binomial and its conjugate follows the difference of squares formula: \((x-y)(x+y) = x^2 - y^2\). This eliminates the square roots in the denominator because \((\sqrt{a})^2 = a\).

In our problem, the denominator was \(\sqrt{2} - 1\). Its conjugate is \(\sqrt{2} + 1\). Multiplying by the conjugate:

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

This is a common technique in algebra and trigonometry when dealing with expressions involving radicals in the denominator.