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