Question

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

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

Answer 1

Answer: (D)

$\tan A-\tan B=x$
$\Rightarrow \frac{1}{\cot A}-\frac{1}{\cot B}=x$
$\Rightarrow \frac{\cot B-\cot A}{\cot A\cot B}=x$
$\Rightarrow \cot A\cdot \cot B=\frac{y}{x}$
$\therefore \cot (A-B)=\frac{\cot A\cot B+1}{\cot B-\cot A}$
$=\frac{\frac{y}{x}+1}{y}=\frac{x+y}{xy}$
$=\frac{1}{x}+\frac{1}{y}$