Understanding the Trigonometry Problem

The question asks us to find the value of cot (A - B) given two relationships between trigonometric functions of angles A and B: tan A - tan B = x and cot B - cot A = y. We need to express the result in terms of x and y.

Let's start by working with the given equations and the expression we need to find.

We are given:

• Equation 1: \tan A - \tan B = x
• Equation 2: \cot B - \cot A = y

We need to find: \cot (A - B)

Using Trigonometric Identities

To solve this, we will use some fundamental trigonometric identities. The key identities are:

• \cot \theta = \frac{1}{\tan \theta}
• \tan (A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}
• \cot (A - B) = \frac{1}{\tan (A - B)} (or using the cot formula directly, but using the tan one and inverting is often simpler when working with tan values)

Step-by-Step Solution

Let's express \cot (A - B) in terms of tangent:

\cot (A - B) = \frac{1}{\tan (A - B)}

Using the tangent subtraction formula:

\cot (A - B) = \frac{1}{\frac{\tan A - \tan B}{1 + \tan A \tan B}}

Inverting the fraction:

\cot (A - B) = \frac{1 + \tan A \tan B}{\tan A - \tan B}

We can split this fraction into two terms:

\cot (A - B) = \frac{1}{\tan A - \tan B} + \frac{\tan A \tan B}{\tan A - \tan B}

From Equation 1, we know that \tan A - \tan B = x . Substitute this into the expression:

\cot (A - B) = \frac{1}{x} + \frac{\tan A \tan B}{x}

Now, let's use Equation 2: \cot B - \cot A = y . We can rewrite this in terms of tangent:

\frac{1}{\tan B} - \frac{1}{\tan A} = y

Find a common denominator on the left side:

\frac{\tan A - \tan B}{\tan A \tan B} = y

We already know that \tan A - \tan B = x from Equation 1. Substitute this into the transformed Equation 2:

\frac{x}{\tan A \tan B} = y

We can rearrange this equation to find the value of \tan A \tan B in terms of x and y:

\tan A \tan B = \frac{x}{y}

Now substitute this value of \tan A \tan B back into the expression for \cot (A - B) we derived earlier:

\cot (A - B) = \frac{1}{x} + \frac{\tan A \tan B}{x}

\cot (A - B) = \frac{1}{x} + \frac{\frac{x}{y}}{x}

Simplify the second term:

\frac{\frac{x}{y}}{x} = \frac{x}{y} \times \frac{1}{x} = \frac{1}{y}

So, substituting this back into the expression for \cot (A - B) :

\cot (A - B) = \frac{1}{x} + \frac{1}{y}

Thus, the value of \cot (A - B) is \frac{1}{x} + \frac{1}{y} .

Conclusion

By using the given equations and applying trigonometric identities, we successfully expressed \cot (A - B) in terms of x and y. The steps involved transforming the second given equation into terms of tangent and using the tangent subtraction formula to expand \cot (A - B) .

Revision Table: Trigonometric Formulas

Additional Information: Connecting tan and cot

Understanding the relationship between tangent and cotangent is crucial in solving problems like this. They are reciprocal functions. This means:

• If you know the value of \tan \theta , you can find \cot \theta by taking its reciprocal ( \cot \theta = \frac{1}{\tan \theta} ), provided \tan \theta \neq 0 .
• Similarly, if you know the value of \cot \theta , you can find \tan \theta by taking its reciprocal ( \tan \theta = \frac{1}{\cot \theta} ), provided \cot \theta \neq 0 .

This reciprocal relationship allowed us to convert the given equation \cot B - \cot A = y into a form involving tangent, \frac{1}{\tan B} - \frac{1}{\tan A} = y , which simplified to \frac{\tan A - \tan B}{\tan A \tan B} = y . This transformation was key to finding the value of \tan A \tan B needed in the solution.