Question

Prove:
sec^2x + tan^2x = (1-sin^4x)sec^4x

Answer 1

what have you tried ?

Answer 2

start from right...
(1-sin^4x)sec^4x = sec^2x (sec^2x - sec^2x sin^4x)

Answer 3

whats next ??

Answer 4

sec^2x (sec^2x - sec^2x sin^4x) = sec^2x (1+tan^2x - (1/cos^2x) sin^4x)

Answer 5

getting too complicated ??

Answer 6

then convert everything into sin and cos...mostly works...

Answer 7

yes the power 4 is confusing

Answer 8

the next step is l.c.m ??

Answer 9

would you like me to continue or convert everything to sin cos ?

Answer 10

no , not sin cos continue the way it is

Answer 11

sec^2x (1+tan^2x - (1/cos^2x) sin^4x)
= sec^2x (1+tan^2x - (sin^2 x/cos^2x) sin^2x)

Answer 12

i need to find an easier way...

Answer 13

got it

Answer 14

:)

Answer 15

(1-sin^4x)sec^4x
= (sec^4x - sin^4 x sec^4 x)
= (sec^4 x - tan^4 x)

Answer 16

=(sec^2 x+ tan^2 x)(sec^2x - tan^2x)
ans whats sec^2x - tan^2x =..?

Answer 17

identities used are \(\sin x*\sec x= \sin x/\cos x = \tan x\)
and \(a^2-b^2=(a+b)(a-b)\)

Answer 18

1

Answer 19

which step you need explaning ? :P

Answer 20

nothing i got it, it was easy

Answer 21

glad to hear :)

Answer 22

\[\frac{ \cos^3x-\sin^3x }{ cosx-sinx }=1+sinx * cosx\]

Answer 23

^ can u tell me start of this prove

Answer 24

\(\large (a^3-b^3)=(a-b)(a^2+ab+b^2)\)

Answer 25

@koli123able

Answer 26

and \(\sin^2x+\cos^2x=1\)

Answer 27

thanks u r the best !!!

Answer 28

not really, but i'll take the compliment :P :)