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Q.

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

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a

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

b

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

c

$\frac{1}{x y}$

d

$x + y$

answer is A.

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Detailed Solution

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

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

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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