Question

tan A+tanB+tanA tanB=1,then A+B=

Answer 1

Answer:

principal value of (A+B)=π/4

General solution of (A+B)=nπ+(π/4) where n∈Z

Step-by-step explanation:

Consider tan(A+B)= tanA +tanB÷(1-tanAtanB).....(1)

Given ,tanA+tanB+tanAtanB=1

tanA+tanB=1-tanAtanB

substitute this in (1)

tan(A+B)=1-tanAtanB÷1-tanAtanB

tan(A+B)=1

therefore A+B=nπ+(π/4) where n∈Z

for principal value n=0 i.e A+B=π/4.