Asked by nani
how do you find the exact value tan(x+y), sinx = (3/5) and secy = (-13/5)
Answers
Answered by
oobleck
sinx = 3/5 so tanx = 3/4 (assuming x is in QI, but it could also be in QII, in which case tanx = -3/4)
secy = -13/5 so tany = -12/5 (assuming y is in QII, But it could also be in QIII, in which case tany = 12/5)
tan(x+y) = (tanx + tany) / (1 - tanx tany)
secy = -13/5 so tany = -12/5 (assuming y is in QII, But it could also be in QIII, in which case tany = 12/5)
tan(x+y) = (tanx + tany) / (1 - tanx tany)
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