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

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

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Answer 1 · 2016-02-20T06:08:00

$sec (x + y) - cos (x + y) = \frac{{(sin x cos y + cos x sin y)}^{2}}{cos x cos y - sin x sin y}$ $sec(x+y)-cos(x+y)=(sin x cos y+cos x sin y)^2/(cos x cos y - sin x sin y)$

Explanation:

Do a little transformation

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

$\frac{1}{cos (x + y)} - cos (x + y)$ $1/cos(x+y)-cos(x+y)$

Use the LCD $cos (x + y)$ $cos (x+y)$

$\frac{1 - {cos}^{2} (x + y)}{cos (x + y)}$ $(1-cos^2(x+y))/cos(x+y)$

From Pythagorean Relation

${sin}^{2} (x + y) = 1 - {cos}^{2} (x + y)$ $sin^2 (x+y)=1-cos^2(x+y)$

so that

$\frac{1 - {cos}^{2} (x + y)}{cos (x + y)}$ $(1-cos^2(x+y))/cos(x+y)$

$\frac{{sin}^{2} (x + y)}{cos (x + y)}$ $(sin^2 (x+y))/cos(x+y)$

From Sum and Difference Formulas

$\frac{{(sin x cos y + cos x sin y)}^{2}}{cos x cos y - sin x sin y}$ $(sin x cos y+ cos x sin y)^2/(cos x cos y-sin x siny)$

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

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Explanation:

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

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