How do you verify the identity $sin(A+B)=(tanA+tanB)/(secAsecB)$ ?

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Answer 1 · 2016-12-01T09:04:56

$sin(A+B)=sinAcosB+cosAsinB=(tanA+tanB)/(secAsecB)$

You know that

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

and, since

$tanalpha=sinalpha/cosalpha$

and $secalpha=1/cosalpha$ ,

then

$(tanA+tanB)/(secAsecB)=$

$=(sinA/cosA+sinB/cosB)/(1/cosA*1/cosB)$

$=((sinAcosB+sinBcosA)/(cancel(cosA)cancelcosB))/(1/(cancelcosAcancelcosB))$

$sinAcosB+sinBcosA$

How do you verify the identity sin(A+B)=(tanA+tanB)/(secAsecB)sin(A+B)=(tanA+tanB)/(secAsecB)?