Question

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

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

Answer 1

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

$\tan A-\tan B=x$ ...(i)
$\cot B-\cot A=y$ ...(ii)
Hence from (ii), we have
$\frac{1}{\tan B}-\frac{1}{\tan A}=y$
$\frac{\tan A-\tan B}{\tan A\tan B}=y$
$\frac{x}{y}=\tan A\tan B$ ... (iii) [from(i)]
Hence, $\cot (A-B)$
$=\frac{1}{\tan (A-B)}$
$=\frac{1+\tan A\tan B}{\tan A-\tan B}$
$=\frac{1+\frac{x}{y}}{x}$
$=\frac{x+y}{xy}$
$=\frac{1}{y}+\frac{1}{x}$
Hence answer is option D