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$\tan A - \tan B = x$ and $\cot B - \cot A = 4$ , then $\cot (A - B) =$

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

\frac{1}{x} - \frac{1}{y}

\frac{1}{x y}

x + y

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detailed solution

Correct option is A

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

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

Now,

$\cot (A - B) = \frac{1}{\tan (A - B)}$

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

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tan⁡A−tan⁡B=xand cot⁡B−cot⁡A=4, then cot⁡(A−B)=

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$\tan A - \tan B = x$ and $\cot B - \cot A = 4$ , then $\cot (A - B) =$