How do you simplify #cos(x+y)/sin(x-y)-2cot(x+y)# to trigonometric functions of x and y?

1 Answer
May 23, 2016

#cos(x+y)/sin(x-y)-2cot(x+y)=1/(sin^2x-sin^2y){sinycosy(3-2sin^2x)-sinxcosx(3-2cos^2y)}#

Explanation:

#cos(x+y)/sin(x-y)-2cot(x+y)#

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

= #cos(x+y)[1/sin(x-y)-2/sin(x+y)]#

= #cos(x+y)[(sin(x+y)-2sin(x-y))/(sin(x-y)sin(x+y))]#

= #cos(x+y)[(sinxcosy+cosxsiny)-2(sinxcosy-cosxsiny))/((sinxcosy+cosxsiny)(sinxcosy+cosxsiny))#

= #cos(x+y){(-sinxcosy+3cosxsiny)/(sin^2xcos^2y-cos^2xsin^2y)}#

= #(cosxcosy-sinxsiny)(-sinxcosy+3cosxsiny)/(sin^2x(1-sin^2y)-(1-sin^2x)sin^2y)#

= #(cosxcosy-sinxsiny)(-sinxcosy+3cosxsiny)/(sin^2x-sin^2y)#

= #1/(sin^2x-sin^2y){(cosxcosy-sinxsiny)(-sinxcosy+3cosxsiny)}#

= #1/(sin^2x-sin^2y){-sinxcosxcos^2y+3cos^2xsinycosy+sin^2xsinycosy-3sinxcosxsin^2y}#

= #1/(sin^2x-sin^2y){-sinxcosx(cos^2y+3sin^2y)+sinycosy(3cos^2x+sin^2x)}#

= #1/(sin^2x-sin^2y){sinycosy(3-2sin^2x)-sinxcosx(3-2cos^2y)}#