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If $\tan A-\tan B=x$ and $\cot B-\cot A=y$, then $\cot (A-B)=$
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2817 Upvotes
Verified Answer
The correct answer is:
$\frac{1}{x}+\frac{1}{y}$
We have,
$$
\begin{aligned}
\tan A-\tan B &=x \\
\text { and } \cot B-\cot A &=y \\
\Rightarrow \frac{1}{\tan B}-\frac{1}{\tan A} &=y \\
\Rightarrow \quad \frac{\tan A-\tan B}{\tan A \tan B} &=y \\
\Rightarrow \quad \tan A \tan B &=\frac{x}{y} \\
\Rightarrow \quad \cot A \cot B &=\frac{y}{x} \\
\text { Now, } \cot (A-B) &=\frac{1+\cot A \cot B}{\cot B-\cot A} \\
&=\frac{1+\frac{y}{x}}{y}=\frac{x+y}{x y}=\frac{1}{y}+\frac{1}{x}
\end{aligned}
$$
$$
\begin{aligned}
\tan A-\tan B &=x \\
\text { and } \cot B-\cot A &=y \\
\Rightarrow \frac{1}{\tan B}-\frac{1}{\tan A} &=y \\
\Rightarrow \quad \frac{\tan A-\tan B}{\tan A \tan B} &=y \\
\Rightarrow \quad \tan A \tan B &=\frac{x}{y} \\
\Rightarrow \quad \cot A \cot B &=\frac{y}{x} \\
\text { Now, } \cot (A-B) &=\frac{1+\cot A \cot B}{\cot B-\cot A} \\
&=\frac{1+\frac{y}{x}}{y}=\frac{x+y}{x y}=\frac{1}{y}+\frac{1}{x}
\end{aligned}
$$
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