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

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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!

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