If tan A = frac{n}{n+1} and tan B = frac{1}{2 | vedclass

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(B) We use the trigonometric identity: $	an (A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$.Substitute the given values: $	an (A+B) = \frac{\frac

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(B) We use the trigonometric identity: $	an (A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$.Substitute the given values: $	an (A+B) = \frac{\frac{n}{n+1} + \frac{1}{2n+1}}{1 - (\frac{n}{n+1} 	imes \frac{1}{2n+1})}$.Simplify the numerator: $\frac{n(2n+1) + 1(n+1)}{(n+1)(2n+1)} = \frac{2n^2 + n + n + 1}{(n+1)(2n+1)} = \frac{2n^2 + 2n + 1}{(n+1)(2n+1)}$.Simplify the denominator: $1 - \frac{n}{(n+1)(2n+1)} = \frac{(n+1)(2n+1) - n}{(n+1)(2n+1)} = \frac{2n^2 + 3n + 1 - n}{(n+1)(2n+1)} = \frac{2n^2 + 2n + 1}{(n+1)(2n+1)}$.Dividing the numerator by the denominator: $	an (A+B) = \frac{2n^2 + 2n + 1}{(n+1)(2n+1)} 	imes \frac{(n+1)(2n+1)}{2n^2 + 2n + 1} = 1$.

If $\tan A = \frac{n}{n+1}$ and $\tan B = \frac{1}{2n+1}$,the value of $\tan (A+B) =$

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