How do you simplify #tan(x+y)/sin(x-y)# to trigonometric functions of x and y?

1 Answer
Dec 21, 2016

This can be simplified to #(sinxcosy + cosxsiny)/(cosxsinx - sinycosy)#

Explanation:

Rewrite #tan(x + y)# as #sin(x + y)/cos(x + y)#.

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

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

We now expand using the formulae #sin(A + B) = sinAcosB + cosAsinB#, #sin(A - B) = sinAcosB - cosAsinB# and #cos(A + B) = cosAcosB - sinAsinB#.

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

#=(sinxcosy + cosxsiny)/((cosxsinxcos^2y - sin^2xsinycosy + cosxsinxsin^2y - cos^2xcosysiny)#

Rearrange in order to look for a factorization in the denominator:

#=(sinxcosy + cosxsiny)/((cosxsinxcos^2y + cosxsinxsin^2y - sin^2xsinycosy - cos^2xcosysiny)#

#=(sinxcosy + cosxsiny)/(cosxsinx(cos^2y + sin^2y) - sinycosy(sin^2x + cos^2x))#

Recall that #sin^2theta + cos^2theta = 1#:

#=(sinxcosy + cosxsiny)/(cosxsinx - sinycosy)#

This is as far as we can go.

Hopefully this helps!