Question

If $\tan A-\tan B=x$ and $\cot B-\cot A=y$, then the true statement is

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

Answer 1

The correct option is A $\cot (A-B)=\frac{1}{x}+\frac{1}{y}$

$\tan A-\tan B=x$, & $\cot B-\cot A=y$ $\Rightarrow \frac{\tan A-\tan B}{\tan A\tan B}=y$

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

Now $\cot (A-B)=\frac{1+\tan A\tan B}{\tan A-\tan B}=\frac{1+\frac{x}{y}}{x}=\frac{1}{x}+\frac{1}{y}$

option A