Question

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

Answer 1

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

Explanation:

`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)`