Question

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

• A. $\frac{1}{y}-\frac{1}{x}$
• B. $\frac{1}{x}-\frac{1}{y}$
• C. $\frac{1}{x}+\frac{1}{y}$
• D. None of the above

Answer 1

Answer: (C)

$\frac{1}{x}+\frac{1}{y}$

$tanA-tanB=x$

$cotB-cotA=y$

$\Rightarrow \frac{1}{tanB}-\frac{1}{tanA}=y$

$\Rightarrow \frac{tanA-tanB}{tanA.tanB}=y$

$\Rightarrow tanAtanB=\frac{x}{y}$

$\Rightarrow cotAcotB=\frac{y}{x}$

$\cot (A-B)=\frac{1+cotAcotB}{cotB-cotA}=\frac{1+\frac{y}{x}}{y}=\frac{x+y}{xy}=\frac{1}{y}+\frac{1}{x}$