:trigonometric identity question

[Edgar92]

URGENT:trigonometric identity question

Homework Statement

tan2x+cos2x+sin2x=sec2x
*the 2 stands for squared since I don't know how to make the squared symbol appear on a compter

Homework Equations

http://www.analyzemath.com/Trigonometry_2/Trigonometric_identities.html
stuff from this website and more

The Attempt at a Solution

(sin2x+cos2x(cos2x)+sin2xcos2x)/cos2x

1+cos2x+sin2xcos2x/cos2x

1+sin2xcos2x

any help is appreciated as I have been at this for over an hour and can't figure it out and it needs to be done soon

 

[rock.freak667]

Homework Equations

http://www.analyzemath.com/Trigonometry_2/Trigonometric_identities.html
stuff from this website and more

The first set of equations shown would help you greatly

tan²x+cos²x+sin²x=sec²x

 

[icystrike]

take note of this identities:
tan²x=sec²x-1
cos²x+sin²x=1

 

[Fightfish]

take note of this identities:
tan²x=sec²x-1
cos²x+sin²x=1

Or you can work from "base principles" replacing sin x with opposite over hypotenuse, cos x with adjacent over hypotenuse et cetera

 

[Mark44]

Or you can work from "base principles" replacing sin x with opposite over hypotenuse, cos x with adjacent over hypotenuse et cetera

A better idea would be to replace tan²(x) with (sin²(x))/(cos²(x)) and sec²(x) with 1/(cos²(x)). That way all your quantities would be in terms of powers of sinx and cosx.