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

Question

1557 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-16T15:16:11

#cos(x-y)/sin(x+y)-cot(x-y)=-(2cos^2xsinycosy+2cosxsinxsin^2y)/(sin^2x-sin^2y)#

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

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

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

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

= #(cos(x-y)(sinxcosy-cosxsiny-sinxcosy-cosxsiny))/(sin^2xcos^2y-cos^2xsin^2y)#

= #(cos(x-y)(-2cosxsiny))/(sin^2x(1-sin^2y)-(1-sin^2x)sin^2y)#

= #-(2cosxsiny(cosxcosy+sinxsiny))/(sin^2x-sin^2xsin^2y-sin^2y+sin^2xsin^2y)#

= #-(2cos^2xsinycosy+2cosxsinxsin^2y)/(sin^2x-sin^2y)##-(2cos^2xsinycosy+2cosxsinxsin^2y)/(sin^2x-sin^2y)#