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