Question

Prove that sec^2xtan^2x +sec^2x= sec^4x

Answer 1

\[\sec^2x \tan^2x+\sec^2x=\sec^4x\]

Answer 2

change sec^2 to its inverse identity, then change tan^2x to its quotient identity first

Answer 3

\[\frac{ 1 }{ \cos^2} * \frac{ \sin^2 }{ \cos^2 }\]

Answer 4

multiply those together

Answer 5

and you get \[\frac{ \sin^2 }{ \cos^4 }\]

Answer 6

Okay I'm following you so far

Answer 7

Then you add the sin^2x/cos^4x to 1/cos^2 ?

Answer 8

ok so then you change sec^2 x to its inverse identity and multiply it by cos^2x on top and bottom to get a common denominator

Answer 9

\[\frac{ 1 }{ \cos^2x}*\frac{ \cos^2x }{ \cos^2x }=\frac{ \cos^2x }{ \cos^4x }\]

Answer 10

And you add in sin so its sin^2x+cosx^2/cos^4 correct?

Answer 11

yep!! from there on the top you look and see that we have a pythagorian identity of sin^2x+cos^2x=1

Answer 12

so we can change the top of the equation to just be 1

Answer 13

does that make sense?

Answer 14

Yup! so youre left with 1/cos^4, which using the inverse identity becomes sec^4x right?

Answer 15

exactly! then you're done (:

Answer 16

Thank you so much for your help!!

Answer 17

no problem!