Question

verify (cosx+sinx)/(cosx-sinx)=sec2x+tan2x

Answer 1

please!

Answer 2

this didn't fall out like I thought it was gonna, sorry. 1 sec....

Above though, since sec^2 = 1/cos^2 and tan^2 = sin^2 / soc^2 you can rewrite it like that

Answer 3

but, its not sec^2 or tan^2x its sec2x and tan2x, they are double angles

Answer 4

well then ignore me completely :P

Answer 5

haha alright. Can you still help me?

Answer 6

yeah, it might make it easier. 1 sec

\[ \frac{\cos x + \sin x}{\cos x - \sin x} = \frac{1+\sin 2x}{\cos 2x}\]

Answer 7

could you explain how you got 1+sin2x/cos2x?

Answer 8

yah
\[ \sec2 x = \frac{1}{\cos2 x} ; \ \ \tan2x = \frac{\sin2x}{\cos2x}\]

they have a common denominator

Answer 9

phew, ok

Answer 10

you'll need double angle identities
\[ \sin 2x = 2 \sin x \cos x
\\ \cos 2x = \cos^2 x - \sin^2 x\]

Then cross multiply

\[ \cos 2x ( \cos x + \sin x) = (1+ \sin 2x)(\cos x - \sin x)\]
and simplify. 6 lines or so. You should end up with 1=1

Have a go at it and see if you get caught up anywhere ^_^

Answer 11

Thanks so much!

Answer 12

^_^