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

Question

1540 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-20T06:08:00

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

Do a little transformation

#sec(x+y)-cos(x+y)#

#1/cos(x+y)-cos(x+y)#

Use the LCD #cos (x+y)#

#(1-cos^2(x+y))/cos(x+y)#

From Pythagorean Relation

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

so that

#(1-cos^2(x+y))/cos(x+y)#

#(sin^2 (x+y))/cos(x+y)#

From Sum and Difference Formulas

#(sin x cos y+ cos x sin y)^2/(cos x cos y-sin x siny)#

have a nice day!