tana = 1/n+1 and tanB = 1/2n+1 than prove that tan (A+B) =1

There is a mistake in the question. tanA=n/n+1tan (A+B)=(tanA +tanB) /1-tanAtanBtan(A+B)={(n/n+1) +(1/2n+1)}/{1-(n/n+1)(1/2n+1)}Take L.C.M.{n(2n+1)+n+1}/(n+1)(2n+1)}/[{(2n+1)(n+1)- n}]/[(n+1)(2n+1)](n+1)(2n+1)are both in the denominators and cancel with each other. tan(A+B)={n(2n+1)+n+1}/{(2n+1)(n+1)-1}=(2n^2+n+n+1)/(2n^2 + n+2n+1-n)=(2n^2 +2n +1)/(2n^2 +2n +1)=1