If Tan A=a/(a+1) and Tan B=1/(2a+1), find the value of (A+B).

tan (A+B) =tanA+tanB/1-tanA.tanB =[(a/a+1)+(1/2a+1)]/[1-(a/a+1)(1/2a+1)]
Simplify the algebra to C as a function of a by multiplying throughout by (a+1)(2a+1)
whence A+B =tan^-1 C

TanA=a/(a+1) TanB=1/(2a+1)

Tan (A+B)=[tan(A) + tan(B)] / [1 - tan(A)tan(B)] Trig Identity
Tan (A+B)= [a/(a+1) + 1/(2a+1)] / [1 -a/(a+1)1/(2a+1)]
Tan(A+B)=[a(2a+1) + (a+1)]/(2a+1)(a+1) / [1 -a/(a+1)(2a+1)]
Tan(A+B)=[2a^2+2a+2]/(2a+1)(a+1)/[(a+1)(2a+1)-a]/(2a+1)(a+1)
Tan (A+B)=[2a^2+2a+2]/[(a+1)(2a+1)-a]=[2a^2+2a+2]/[2a^2+3a+1-a]
Tan(A+B)=[2a^2+2a+2]/[2a^2+2a+1]=1
Tan^-1(1)=A+B
A+B=pi/4

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