Question

verify

secˆ2y+tanˆ2y=(1−sinˆ4y)secˆ4y

I got as far as. Cross multiplying between the 2 sides is not allowed.

Answer 1

\[\secˆ2y+\tanˆ2y=(1−\sinˆ4y)\secˆ4y\]

Answer 2

\[\frac{ 1+\sin ^{2} y}{\cos ^{2} y }=\frac{ 1-\sin ^{4}y }{ \cos ^{4}y }\]

Answer 3

factorize the numerator in the right side and simplify.., then you'll get it equals to the left side...

Answer 4

\[\sec^2 x+\tan^2 x=1∗(\sec^2 x+\tan^2 x)=(\sec^2 x−\tan^2 x)∗(\sec^2 x+\tan^2 x)=\]

Answer 5

the identity you use is not the right one.

Answer 6

\[\sec^4 x - \tan^4 x=(1 - \sin^4 x)\sec^4x\]

Answer 7

the identity is right
1+tan^2x=sec^2x
i put wrong equation above

Answer 8

that is what i was to tell you

Answer 9

thanks for the correction :)

Answer 10

how did you come up with this?

1∗(sec2x+tan2x)=(sec2x−tan2x)∗(sec2x+tan2x)

Answer 11

\[(1-\sin^4 y)(\sec^4y)=(1-\sin^2y)(1+\sin^2y)\sec^4y=\cos^2y(1+\sin^2y)\sec^4y\]
\[\cos^2y \sec^4y(1+\sin^2y)=\sec^2y(1+\sin^2y)\]
\[\sec^2y+\sec^2y \sin^2y=\sec^2y+\frac{ \sin^2y }{\cos^2y }=\sec^2y+\tan^2y\]

Answer 12

it is verified

Answer 13

cos^2x+sin^2x=1
dividing by cos^2x,
1+tan^2x=sec^2x
thus sec^2-tan^2=1

Answer 14

got it! thank you all for you help!