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How do you simplify #4cos(x+y)-tan(x-y)/cos(x+y)# to trigonometric functions of x and y?

1 Answer

Shwetank Mauria

Dec 24, 2016

#4cos(x+y)-tan(x-y)/cos(x+y)=4cosxcosy-4sinxsiny-(sinxcosy-cosxsiny)/(cos^2x-sin^2y)#

Explanation:

#4cos(x+y)-tan(x-y)/cos(x+y)#

= #4(cosxcosy-sinxsiny)-sin(x-y)/(cos(x-y)cos(x+y))#

= #4cosxcosy-4sinxsiny-(sinxcosy-cosxsiny)/(cos^2xcos^2y-sin^2xsin^2y)#

= #4cosxcosy-4sinxsiny-(sinxcosy-cosxsiny)/((1-sin^2x)(1-sin^2y)-sin^2xsin^2y)#

= #4cosxcosy-4sinxsiny-(sinxcosy-cosxsiny)/(1-sin^2x-sin^2y)#

= #4cosxcosy-4sinxsiny-(sinxcosy-cosxsiny)/(cos^2x-sin^2y)#

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