If tan A - tan B = x and cot A - cot B = y,th | vedclass

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(D) Given: $	an A - 	an B = x$ and $\cot A - \cot B = y$. We know that $	an A - 	an B = \frac{1}{\cot A} - \frac{1}{\cot B} = \frac{\cot B - \cot

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

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(D) Given: $	an A - 	an B = x$ and $\cot A - \cot B = y$. We know that $	an A - 	an B = \frac{1}{\cot A} - \frac{1}{\cot B} = \frac{\cot B - \cot A}{\cot A \cot B} = x$. Since $\cot A - \cot B = y$,we have $\cot B - \cot A = -y$. Substituting this into the equation: $\frac{-y}{\cot A \cot B} = x$,which implies $\cot A \cot B = -\frac{y}{x}$. Now,using the formula $\cot (A - B) = \frac{\cot A \cot B + 1}{\cot B - \cot A}$: $\cot (A - B) = \frac{-\frac{y}{x} + 1}{-y} = \frac{\frac{x - y}{x}}{-y} = \frac{y - x}{xy}$. Thus,the correct option is $D$.

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

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