trig identities help !

**mike--1988** · December 28th 2009, 02:07 AM

hi everyone

im ok with normal trig identiies but im confudes with theres any help please :

heres the question :

1) sin^4x - cos^4x = sin^2x - cos^2x

2)sec^4x - tan^4x = 1 + 2tan^2x

...thanks

**Prove It** · December 28th 2009, 02:21 AM

Originally Posted by mike--1988

hi everyone

im ok with normal trig identiies but im confudes with theres any help please :

heres the question :

1) sin^4x - cos^4x = sin^2x - cos^2x

2)sec^4x - tan^4x = 1 + 2tan^2x

...thanks

1) $\sin^4{x} - \cos^4{x} = (\sin^2{x} + \cos^2{x})(\sin^2{x} - \cos^2{x})$

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

$= \sin^2{x} - \cos^2{x}$ .

2) $\sec^4{x} - \tan^4{x} = (\sec^2{x})^2 - \tan^4{x}$

$= (1 + \tan^2{x})^2 - \tan^4{x}$

$= 1 + 2\tan^2{x} + \tan^4{x} - \tan^4{x}$

$= 1 + 2\tan^2{x}$ .

**Soroban** · December 28th 2009, 04:39 AM

Hello, mike--1988!

Another approach to #2 . . .

$2)\;\sec^4\!x - \tan^4\!x \;=\; 1 + 2\tan^2\!x$

$\text{Factor: }\:\sec^4\!x - \tan^4\!x \:=\:(\sec^2\!x + \tan^2\!x)\underbrace{(\sec^2\!x - \tan^2\!x)}_{\text{This is 1}}$
. . . . . . . . . . . . . . $=\;\sec^2\!x + \tan^2\!x$

. . . . . . . . . . . . . . $=\;(\tan^2\!x + 1) + \tan^2\!x$

. . . . . . . . . . . . . . $= \;1 + 2\tan^2\!x$

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