Question 196593
Since {{{tan(x)=sin(x)/cos(x)}}}, we can say 



{{{tan(A+B)=sin(A+B)/cos(A+B)}}}


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{{{tan(A+B)=sin(A+B)/cos(A+B)}}} Start with the given equation.



{{{tan(A+B)=(sin(A)cos(B)+cos(A)sin(B))/cos(A+B)}}} Expand sine using the sum difference identity



{{{tan(A+B)=(sin(A)cos(B)+cos(A)sin(B))/(cos(A)cos(B)-sin(A)sin(B))}}} Expand cosine using the sum difference identity



{{{tan(A+B)=((sin(A)cos(B))/cos(A)+(cos(A)sin(B))/cos(A))/((cos(A)cos(B))/cos(A)-(sin(A)sin(B))/cos(A))}}} Divide EVERY term by {{{cos(A)}}}



{{{tan(A+B)=(tan(A)cos(B)+sin(B))/(cos(B)-tan(A)sin(B))}}} Reduce and simplify



{{{tan(A+B)=((tan(A)cos(B))/cos(B)+sin(B)/cos(B))/(cos(B)/cos(B)-(tan(A)sin(B))/cos(B))}}} Divide EVERY term by {{{cos(B)}}}



{{{tan(A+B)=(tan(A)+tan(B))/(1-tan(A)tan(B))}}} Reduce and simplify




So this verifies {{{tan(A+B)=(tan(A)+tan(B))/(1-tan(A)tan(B))}}}