Question

If $\tan A-\tan B=x$ and $\cot B-\cot A=y$, where $x,y\ne 0$, then $\cot (A-B)$ is

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

Answer 1

The correct option is B $\frac{1}{x}+\frac{1}{y}$

We have $\tan A-\tan B=x$ and $\cot B-\cot A=y$
Now, $y=\frac{\tan A-\tan B}{\tan A\tan B}$
$\Rightarrow y=\frac{x}{\tan A\tan B}\Rightarrow \cot A\cot B=\frac{y}{x}$
Now,
$\cot (A-B)=\frac{\cot A\cot B+1}{\cot B-\cot A}$
$\cot (A-B)=\frac{\frac{y}{x}+1}{y}=\frac{x+y}{xy}=\frac{1}{x}+\frac{1}{y}$