Science

Math

Humanities

... and beyond

Socratic Meta
Featured Answers

Topics

Question #42bd5

1 Answer

Nov 28, 2016

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

$= \frac{1 - sin x}{(1 - sin x) (1 + sin x)} + \frac{sin x}{{cos}^{2} x}$ $=(1-sinx)/((1-sinx)(1+sinx))+sinx/cos^2x$

$= \frac{1 - sin x}{1 - {sin}^{2} x} + \frac{sin x}{{cos}^{2} x}$ $=(1-sinx)/(1-sin^2x)+sinx/cos^2x$

$= \frac{1 - sin x}{{cos}^{2} x} + \frac{sin x}{{cos}^{2} x}$ $=(1-sinx)/cos^2x+sinx/cos^2x$

$= \frac{1 - sin x + sin x}{{cos}^{2} x}$ $=(1-sinx+sinx)/cos^2x$

$= \frac{1}{{cos}^{2} x} = {sec}^{2} x$ $=1/cos^2x=sec^2x$

Related questions

See all questions in Double Angle Identities

Impact of this question

1495 views around the world

You can reuse this answer
Creative Commons License