Verify: #-(cotA+cotB)/(cotA-cotB) = sin(A+B)/sin(A-B)# ?

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Answer 1 · 2016-07-30T01:35:57

Expand and simplify RHS then apply identities for #sin(A+-B)#

RHS =#-(cotA+cotB)/(cotA-cotB) = -(cosA/sinA+cosB/sinB)/(cosA/sinA-cosB/sinB)#

RHS=#-(cosAsinB + cosBsinA)/(sinAsinB)/(cosAsinB-cosBsinA)/(sinAsinB)#

RHS=#-(cosAsinB + cosBsinA)/(cancel(sinAsinB))/(cosAsinB-cosBsinA)/(cancel(sinAsinB))#

RHS=#-(cosAsinB + cosBsinA)/(cosAsinB-cosBsinA)#

#=-sin(A+B)/-sin(A-B) = sin(A+B)/sin(A-B) =# LHS

Answer 2 · 2016-07-30T04:28:24

#LHS=sin(A+B)/sin(A-B)#

#=(sinAcosB+cosAsinB)/(sinAcosB-cosAsinB)#

Dividing both numerator and denominator by #sinAsinB#

#=((sinAcosB)/(SinAsinB)+(cosAsinB)/(sinAsinB))/((sinAcosB)/(sinAsinB)-(cosAsinB)/(sinAsinB))#

#=-((cotA+cotB)/(cotA-cotB))=RHS#

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