Question

help me prove this
plz use only sin^2A+cos^2A=1formet

sec^4A-1=2tan^2A+tan^4A

Answer 1

Well with that format we can't use this for sec and tan

Answer 2

By that I mean, we should use this formula using your formula \[\Large \sin^2+\cos^2=1\]
Divide all sides by cos and you get \[\Large {\sin^2 \over \cos^2} +\frac{\cos^2}{\cos^2}=\frac{1}{\cos^2}\] \[\Large \tan^2+1=\sec^2\]

Answer 3

rearranging we get \[\Large \tan^2=\sec^2-1\]

Answer 4

use 1+tan^2A=sec^2A AND cot^2+1=cosec^2

Answer 5

So if we factor \[\Large (\sec^4-1)=(\sec^2-1)(\sec^2+1)\]

Answer 6

Remembering that \[\Large \sec^2-1 =\tan^2\]

We have \[\Large \tan^2(\sec^2+1)\]

Answer 7

So now replace sec^2 with the identified before and tell me what you get

Answer 8

the identity

Answer 9

okay

Answer 10

(1+tan^2)+(1+tan^2)^4 this okay?

Answer 11

Your notation is incorrect though

Answer 12

\[\Large \tan^2((1+\tan^2)+1)\]

Answer 13

So with that simplified, you can get your other side of the equation

Answer 14

l.h.s
sec^4A-1
=(sec^2A-1)(sec^2A+1)
=tan^2A{(tan^2A+1)+1}



am i okay rng agn?

Answer 15

That's fine, the two 1s add together and then that's where you get the 2tan^2 and the tan^2s multply together and you get tan^4

Answer 16

oh now
=tan^2A(tan^2A+2)
=tan^4a+2tan^2A
=r.hs


i think now okay

Answer 17

thanks you doulikepiecauseidont.

Answer 18

Yw. Anytime