Question

Literally no idea and have been on it for an hour! Prove: (tanx+secx)^2 = 2 secx+2tanxsecx-1

Answer 1

square it !!

Answer 2

Square the right hand side?

Answer 3

then use a well known identity involving secant squared and tangent squared

Answer 4

no, the left hand side has an exponent of two
multiply that out

Answer 5

Well I get tan^2x+2tanxsecx+sec^2x ?

Answer 6

@satellite73

Answer 7

yes

Answer 8

now look for an identity involving secant squared and tangent square
you are stuck with the \(2\tan(x)\sec(x)\)

Answer 9

I know that sec^2x-1=tan^2x

Answer 10

I just am not the best with trigonometry, so I get confused with this sometimes

Answer 11

what happesn if you replace \(\tan^2(x)\) by \(\sec^2(x)-1\)?

Answer 12

Would the secants cancel out and I would be left with 2tanxsecx -1?

Answer 13

Oh wait, no, the secants would add together!

Answer 14

Ohhhhh!

Answer 15

yeah, but you don't actually get what you wrote above, you get \[2\sec^2(x)+2\sec(x)\tan(x)-1\]

Answer 16

That's the right side of the equation, though! So that's all we need to do?

Answer 17

it is ? you didn't put a square over the secant, but if it i supposed to be there then we are done

Answer 18

in fact it has to be there, so maybe it was just a typo

Answer 19

Oh sorry, yes... I am looking at my paper and it is (tanx+secx)^2 = 2 sec^2x+2tanxsecx-1

Answer 20

Awesome! Thanks so much :)

Answer 21

yw
now too bad was it (certainly not worth an hour of your time!)

Answer 22

Haha not at all! Some proofs come really easy to me and others for some reason stump me! Lol; I think I overthink it :p