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If $\tan A-\tan B=x$ and $\cot B-\cot A=y$, then what is cot
$(A-B)$ equal to?
Options:
$(A-B)$ equal to?
Solution:
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Verified Answer
The correct answer is:
$\frac{1}{x}+\frac{1}{y}$
Let $\tan A-\tan B=x$ and $\cot B-\cot A=y$.
$\Rightarrow \frac{1}{\tan B}-\frac{1}{\tan A}=y$
$\Rightarrow \frac{\tan A-\tan B}{\tan A \tan B}=y \Rightarrow \frac{x}{\tan A \tan B}=y$
Consider $\cot (A-B)=\left(\frac{1}{\tan (A-B)}\right)$
$=\frac{1+\tan A \tan B}{\tan A-\tan B}=\frac{1+\frac{x}{y}}{x}=\frac{y+x}{x y}=\frac{1}{x}+\frac{1}{y}$.
$\Rightarrow \frac{1}{\tan B}-\frac{1}{\tan A}=y$
$\Rightarrow \frac{\tan A-\tan B}{\tan A \tan B}=y \Rightarrow \frac{x}{\tan A \tan B}=y$
Consider $\cot (A-B)=\left(\frac{1}{\tan (A-B)}\right)$
$=\frac{1+\tan A \tan B}{\tan A-\tan B}=\frac{1+\frac{x}{y}}{x}=\frac{y+x}{x y}=\frac{1}{x}+\frac{1}{y}$.
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