Question

If \( \tan A + \tan B = x \) and \( \cot A + \cot B = y \), then \( \tan(A + B) = \)

• A. \( \frac{xy}{x - y} \)
• B. \( \frac{xy}{y - x} \)
• C. \( \frac{xy}{x + y} \)
• D. \( \frac{x - y}{xy} \)

Hint:

Use the identity for \( \tan(A + B) \) and manipulate the given expressions using trigonometric identities to find the desired result.

Answer 1

The Correct Option is B

We are given the following equations: - \( \tan A + \tan B = x \) - \( \cot A + \cot B = y \) We need to find the value of \( \tan(A + B) \). We use the identity for \( \tan(A + B) \): \[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \] From the given, \( \tan A + \tan B = x \), so we have: \[ \tan(A + B) = \frac{x}{1 - \tan A \tan B} \] Now, we express \( \tan A \tan B \) in terms of \( y \). We know that: \[ \cot A + \cot B = y \quad \Rightarrow \quad \frac{1}{\tan A} + \frac{1}{\tan B} = y \] This can be rewritten as: \[ \frac{\tan A + \tan B}{\tan A \tan B} = y \quad \Rightarrow \quad \frac{x}{\tan A \tan B} = y \] Thus, \[ \tan A \tan B = \frac{x}{y} \] Now substitute this back into the expression for \( \tan(A + B) \): \[ \tan(A + B) = \frac{x}{1 - \frac{x}{y}} = \frac{x}{\frac{y - x}{y}} = \frac{xy}{y - x} \] Thus, the correct answer is option (2).

Answer 2

Given:
\[ \tan A + \tan B = x, \quad \cot A + \cot B = y \]
We are to find: \( \tan(A + B) = ? \)

Step 1: Use tangent addition identity
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} = \frac{x}{1 - \tan A \tan B} \] So we need to find \( \tan A \tan B \)

Step 2: Use cotangent sum identity
Recall: \[ \cot A + \cot B = \frac{1}{\tan A} + \frac{1}{\tan B} = \frac{\tan B + \tan A}{\tan A \tan B} = \frac{x}{\tan A \tan B} \] So: \[ y = \frac{x}{\tan A \tan B} \Rightarrow \tan A \tan B = \frac{x}{y} \]

Step 3: Substitute into tan(A + B)
\[ \tan(A + B) = \frac{x}{1 - \frac{x}{y}} = \frac{x}{\frac{y - x}{y}} = \frac{xy}{y - x} \]

Final Answer:
\[ \boxed{\frac{xy}{y - x}} \]