Question

If $\tan A - \tan B = x$ and $\cot B - \cot A = y$, then the expression for $\cot(A - B)$ is:

• A. $\frac{1}{x-y}$
• B. $\frac{1}{x+y}$
• C. $\frac{1}{x} + y$
• D. $\frac{1}{x} - \frac{1}{y}$
• E. $\frac{1}{x} + \frac{1}{y}$

Hint:

This is a standard identity. $\cot(A-B)$ can be visualized as the sum of the reciprocals of the differences of the tangents and cotangents. If you see $\tan A - \tan B$ and $\cot B - \cot A$, the answer is almost always $1/x + 1/y$.

Answer 1

Answer: (E)

Concept: The goal is to relate the given differences of tangents and cotangents to the formula for $\cot(A-B)$. Since $\cot(A-B) = \frac{1}{\tan(A-B)}$, we should first find an expression for $\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$.

Step 1: Transform the cotangent condition into tangent terms.
The second given condition is $\cot B - \cot A = y$. Using the identity $\cot \theta = 1/\tan \theta$: \[ \frac{1}{\tan B} - \frac{1}{\tan A} = y \] Taking a common denominator $\tan A \tan B$: \[ \frac{\tan A - \tan B}{\tan A \tan B} = y \]

Step 2: Substitute the first condition to find the product $\tan A \tan B$.
We know $\tan A - \tan B = x$. Substitute $x$ into the equation from
Step 1: \[ \frac{x}{\tan A \tan B} = y \quad \Rightarrow \quad \tan A \tan B = \frac{x}{y} \]

Step 3: Calculate $\cot(A - B)$ using the reciprocal of the tangent sum formula.
First, find $\tan(A-B)$: \[ \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} = \frac{x}{1 + \frac{x}{y}} = \frac{x}{\frac{y+x}{y}} = \frac{xy}{x+y} \] Now, the cotangent is the reciprocal: \[ \cot(A - B) = \frac{1}{\tan(A-B)} = \frac{x+y}{xy} \] Splitting the fraction gives the final result: \[ \cot(A - B) = \frac{x}{xy} + \frac{y}{xy} = \frac{1}{y} + \frac{1}{x} \]