# Thread: trig identities help !

1. ## trig identities help !

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

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

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

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

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

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

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

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

3. Hello, mike--1988!

Another approach to #2 . . .

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

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

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

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