Question

If $\tan A-\tan B=x$ & $\cot B-\cot A=y$, then

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

Given $\tan A-\tan B=x$

And also $\cot B-\cot A=y⟹\frac{1}{\tan B}-\frac{1}{\tan A}=y⟹\frac{\tan A-\tan B}{\tan A\tan B}=y⟹\tan A\tan B=\frac{x}{y}$

As we know that

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

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