Prove that $(sin(A+B))/(cos(A-B)) = (tan A-tan B)/(1+tan A tan B)$ ?

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Answer 1 · 2018-04-12T16:30:53

$(tanA-tanB)/(1+tanAtanB)=tan(A-B)=sin(A-B)/cos(A-B)$ .

Check the Problem.

Answer 2 · 2018-04-12T16:30:55

Please, see the proof below

$"Reminder"$

$sin(A+B)=sinAcosB+sinBcosA$

$cos(A-B)=cosAcosB+sinAsinB$

Therefore,

$LHS=sin(A+B)/cos(A-B)$

$=(sinAcosB+sinBcosA)/(cosAcosB+sinAsinB)$

$=((sinAcosB)/(cosAcosB)+(sinBcosA)/(cosAcosB))/((cosAcosB)/(cosAcosB)+(sinAsinB)/(cosacosB))$

$=(tanA+tanB)/(1+tanAtanB)$

$=RHS?$

$QED$

Prove that (sin(A+B))/(cos(A-B)) = (tan A-tan B)/(1+tan A tan B)(sin(A+B))/(cos(A-B)) = (tan A-tan B)/(1+tan A tan B)?