Question

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

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

Answer 1

Answer: (A)

Given $\tan A-\tan B=x...\left(i\right)$
and $\cot B-\cot A=y...\left(ii\right)$

From Eq(ii)
$\cot B-\cot A=y$

we get
$\Rightarrow \frac{1}{\tan B}-\frac{1}{\tan A}=y$

$\Rightarrow \frac{\tan A-\tan B}{\tan B\tan A}=y$

$\Rightarrow \frac{x}{\tan B\tan A}=y$

$\therefore \cot (A-B)=\frac{\cot A\cot B+1}{\cot B-\cot A}$
$=\frac{\frac{y}{x}+1}{y}$............... (Eq(ii) and Eq(iii)
$=\frac{1}{x}+\frac{1}{y}$