Question

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

Answer 1

...is this supposed to be \[\tan(x) - \sec^4(x) = 1-2\sec^2(x)\]?

Answer 2

oh no. oops. its tan^4(x) - sec^4(x) = 1 - 2 sec^2(x)

Answer 3

ok you have the difference of squares here a^2+b^2=(a-b)(a+b)

Answer 4

okay so that would make it (tan^2x-sec^2x) (tan^2x+sec^2x)

Answer 5

so we have (tan^2x)^2-(sec^2x)^2=[(tanx)^2+(secx)^2][(tanx)^2-(secx)^2]

Answer 6

yes and 1=(secx)^2-(tanx)^2

Answer 7

so we have -((tanx)^2+(sec^2))((secx)^2-(tanx)^2)

Answer 8

-((tanx)^2+(secx)^2)(1)

Answer 9

ok the other side is in terms of sec so put this side in terms of sec

Answer 10

ok
thanks

Answer 11

you got it?

Answer 12

this problem is basically done

Answer 13

so we have -((secx)^2-1+(secx)^2)=-(2(secx)^2-1)=1-2(secx)^2

Answer 14

which is the other side of the equation