Question

Prove the identity
(tanx)/(1-cosx) = cscx(1+secx)

Answer 1

I've gotten up to
(sinx/cosx)(1/(1-cosx)

Answer 2

is that correct so far?

Answer 3

ok and you can write that as:\[\frac{\frac{\sin(x)}{\cos(x)}}{1-\cos(x)} \\ \frac{\sin(x)}{\cos(x)} \cdot \frac{1}{1-\cos(x)}\]

Answer 4

oh that is what you wrote

Answer 5

lol

Answer 6

\[\frac{\sin(x)}{\cos(x)} \cdot \frac{1}{1-\cos(x)} \cdot \frac{1+\cos(x)}{1+\cos(x)}\]
see what this does maybe

Answer 7

multiply the second and last fraction

Answer 8

so the last fraction is just like multiplying by 1 right?

Answer 9

yep

Answer 10

It'd be
(sinx/cosx)((1+cosx)/(1-cos^2x))

Answer 11

right and 1-cos^2(x) is sin^2(x)

Answer 12

\[\frac{\sin(x)}{\cos(x)} \cdot \frac{1+\cos(x)}{\sin^2(x)} \\ \frac{\sin(x)}{\sin^2(x)} \frac{1+\cos(x)}{\cos(x)}\]

Answer 13

can you see where to go from here?

Answer 14

Emm no haha sorry.

Answer 15

\[\frac{\sin(x)}{\sin^2(x)}(\frac{1}{\cos(x)}+\frac{\cos(x)}{\cos(x)})\]
how about here?

Answer 16

cos(x)/cos(x)=?
1/cos(x)=?
sin(x)/sin^2(x)=?

Answer 17

1) = 1
2) = secx
3) I'm not sure about that one :P

Answer 18

where did you get the cosx/cosx?

Answer 19

1) good
2) good
3) do you know how to simplify say something like u/u^2 ?

separated the fraction to "where did you get the cosx/cosx?
"
\[\frac{a+b}{c}=\frac{a}{c}+\frac{b}{c} \\ \text{ we had } \frac{1+\cos(x)}{\cos(x)}=\frac{1}{\cos(x)}+\frac{\cos(x)}{\cos(x)}\]

Answer 20

aah I see, ok :)

Answer 21

3) um, i'm not sure.

Answer 22

\[\frac{u}{u^2}=\frac{u}{u \cdot u}=\frac{\cancel{u}}{\cancel{u} u}=\frac{1}{u} \\ \text{ or just use law of exponents } \\ \frac{u}{u^2}=u^{1-2}=u^{-1}=\frac{1}{u} \\ \frac{\sin(x)}{\sin^2(x)}=\frac{1}{\sin(x)}\]

Answer 23

and 1/sin(x) is....

Answer 24

cscx!

Answer 25

yep

Answer 26

What we did:
\[\frac{\tan(x)}{1-\cos(x)} \\ \tan(x) \cdot \frac{1}{1-\cos(x)} \\ \frac{\sin(x)}{\cos(x)} \frac{1}{1-\cos(x)} \\ \frac{\sin(x)}{\cos(x)} (\frac{1}{1-\cos(x) } \cdot 1) \\ \frac{\sin(x)}{\cos(x)} (\frac{1}{1-\cos(x) } \cdot \frac{1+\cos(x)}{1+\cos(x)} ) \\ \frac{\sin(x)}{\cos(x)}(\frac{1+\cos(x)}{1-\cos^2(x)}) \\ \frac{\sin(x)}{\cos(x)} \frac{1+\cos(x)}{\sin^2(x)} \\ \frac{\sin(x)}{\sin^2(x)} \frac{1+\cos(x)}{\cos(x)} \\ \frac{1}{\sin(x)} \frac{1+\cos(x)}{\cos(x)} \\ \frac{1}{\sin(x)} (\frac{1}{\cos(x)}+\frac{\cos(x)}{\cos(x)}) \\ \csc(x)(\sec(x)+1) \\ \csc(x)(1+\sec(x))\]

Answer 27

aw thanks for taking the time to type that all up!

Answer 28

haha I guess I should do a few more practice ones. Sigh. Thanks so much for the help :)

Answer 29

np
trig identities can be tricky (Takes practice practice)
but fun
just think you are doing a puzzle or something
puzzles should be fun

Answer 30

Tricky may be an understatement xP
haha sure sure.