How do you prove # [sin(x+y) / sin(x-y)] = [(tan(x) + tan (y) )/ (tan (x) - tan (y))]#?

We have, #tanx+tany=sinx/cosx+siny/cosy#,

#=(sinxcosy+cosxsiny)/(cosxcosy)#,

#rArr tanx+tany=sin(x+y)/(cosxcosy)...................(square^1)#.

Similarly, #tanx-tany=sin(x-y)/(cosxcosy)............(square^2)#.

#:." by "(square^1) and (square^2), (tanx+tany)/(tanx-tany)#,

#=sin(x+y)/sin(x-y)#.

Hence, the Proof.

We seek to prove that:

# sin(x+y)/sin(x-y) -= (tanx+tany)/(tanx-tany) #

We can the trigonometric identities:

# sin(A+B) -= sinAcosB + cosAsinB #
# sin(A-B) -= sinAcosB - cosAsinB #

Consider the LHS:

# LHS = sin(x+y)/sin(x-y) #

# \ \ \ \ \ \ \ \ = (sinxcosy + cosxsiny)/(sinxcosy - cosxsiny) #

Now if we multiply both numerator and denominator by # 1/(cosxcosy)# we get:

# LHS = (sinxcosy + cosxsiny)/(sinxcosy - cosxsiny) * (1/(cosxcosy)) / (1/(cosxcosy))#

# \ \ \ \ \ \ \ \ = ( (sinxcosy)/(cosxcosy) + (cosxsiny)/(cosxcosy))/ ((sinxcosy)/(cosxcosy) - (cosxsiny)/(cosxcosy)) #

# \ \ \ \ \ \ \ \ = ( (sinx)/(cosx) + (siny)/(cosy))/ ((sinx)/(cosx) - (siny)/(cosy)) #

# \ \ \ \ \ \ \ \ = ( tanx + tany )/ ( tanx - tany ) \ \ \ # QED