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#