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

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

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

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(D) Given: $	an A - 	an B = x$ and $\cot B - \cot A = y.$ We know that $\cot B - \cot A = \frac{1}{	an B} - \frac{1}{	an A} = \frac{	an A - 	an B}{	an A 	an B} = y.$ Substituting $x = 	an A - 	an B$,we get $\frac{x}{	an A 	an B} = y$,which implies $	an A 	an B = \frac{x}{y}.$ Now,$\cot (A - B) = \frac{1}{	an (A - B)} = \frac{1 + 	an A 	an B}{	an A - 	an B}.$ Substituting the values,$\cot (A - B) = \frac{1 + \frac{x}{y}}{x} = \frac{1}{x} + \frac{x}{xy} = \frac{1}{x} + \frac{1}{y}.$

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

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