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

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(B) Given,$	an A + 	an B = x$ and $\cot A + \cot B = y$. Since $\cot A + \cot B = \frac{1}{	an A} + \frac{1}{	an B} = \frac{	an A + 	an B}{	an

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

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(B) Given,$	an A + 	an B = x$ and $\cot A + \cot B = y$. Since $\cot A + \cot B = \frac{1}{	an A} + \frac{1}{	an B} = \frac{	an A + 	an B}{	an A 	an B} = y$. Substituting $	an A + 	an B = x$,we get $\frac{x}{	an A 	an B} = y$,which implies $	an A 	an B = \frac{x}{y}$. Now,using the formula $	an (A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,$	an (A + B) = \frac{x}{1 - \frac{x}{y}} = \frac{x}{\frac{y - x}{y}} = \frac{xy}{y - x}$.

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

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