Question

prove the identity
csc theta over tan theta + cot theta = cos theta

Answer 1

\[\frac{ \frac{1}{sin {\theta}}}{tan{\theta}}+\frac{1}{tan{\theta}}=cos{\theta}\]

Answer 2

I am sorry, how did that happen?

Answer 3

Sorry I will write it up first and post it with steps. Just a moment

Answer 4

Sweet, thanks!

Answer 5

does it mean that (1/sin x)^2+(1/sin x)=1?

Answer 6

if you divide the cos theta from both sides

Answer 7

Yes correct

Answer 8

I still would appreciate to see the steps plz :s

Answer 9

but why? let x=pi/6 then 4+2=1??

Answer 10

yes I'm having that issue too. Are you sure it's written down right?

Answer 11

\[\csc \theta \over \tan \theta +\cot \theta\]
=
[\cos \theta\]

Answer 12

I see now, next time put the tan theta + cot theta in brackets to make it clearer.

Answer 13

multiply the LHS by sinx cosx

Answer 14

gotcha. sorry about that.

Answer 15

i mean csc x /(tan x + cot x)= cscx sinx cosx / (tanx sinx cosx+cotx sinx cosx)=cosx/(sin^2 x+cos^2 x)=cosx

Answer 16

im sorry it should be theta...

Answer 17

why did you multiply each sides by sinx cosx? how could u tell?

Answer 18

\[\frac{1/sin{\theta}}{tan{\theta}+1/tan{\theta}}=cos{\theta}\]\[1/sin{\theta}=tan{\theta}cos{\theta}+cos{\theta}/tan{\theta}\]\[1/sin{\theta}=sin{\theta}+cos^2{\theta}/sin{\theta}\]\[1/sin{\theta}=sin^2{\theta}/sin{\theta}+cos^2{\theta}/sin{\theta}\]\[1/sin{\theta}=1/sin{\theta}\]\[1=1\]

Answer 19

Here's a different way to go about it :)

Answer 20

because there are sinx cosx in the denominator

Answer 21

my teacher doesn't let me combite the left side and the right side.. in this case i would need to make the left side cos theta. :S

Answer 22

zed; yur way is something to think about, and it's cool.
yet i won't b able to use your method

Answer 23

@Zed you shouldnt give out a prove like this, you should write LHS=...=....=...=RHS

Answer 24

yifian; i don't really get how you got there.. :(

Answer 25

i can't figure it out

Answer 26

which step?

Answer 27

is there other way to do it?

Answer 28

Thanks Yifan, I haven't done proof in a long time.

\[LHS = \frac{1/sinx}{sinx/cosx+cosx/sinx}\]
\[=\frac{1/sinx}{(sin^2x+cos^2x)/sinxcosx}\]\[=\frac{1/sinx}{1/sinxcosx}\]\[=\frac{1}{cosx}\]\[=RHS\]

Answer 29

sorry that second last line was supposed to be cosx

Answer 30

yeah that is a good prove except that you substitute theta for x..

Answer 31

yeah i got tired of writing theta out

Answer 32

thanks Yifan

Answer 33

1 over cos x isn't same as cos x

Answer 34

It was a typo

Answer 35

i got that too.

Answer 36

well thanks for your time :)

Answer 37

i still don't get it . for some reason, im very confused. i will ask my teacher and I will let you know if there is a simpler way to do it if you are interested :P g'nite.