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Trig question

13

Been stuck since last night
so I give up plase help me with this

Prove that
Cot2x+cosec2X=cotX
done these
1/(tan2x+sin2x)
1/( (2tanx/(1-tan^2 x)) + 2sinx cosx)
after that I made the domenators(sP?) the same

so that 1/ (2 tanx +2sinxcosx-2sin x cosx tan^2 x)/(1-tan^2 x)
now the 1- tan ^ 2 x becomes the nomenator(sp?)

so (1- tan^2 x)/2tanx + 2sinx cosx -2sinx cosx tan^2 x
after going this far I seem to do something really stupid and get lost between the numbers not knowing how to go back...pleaase help me

Reply 1

elpaw

15

habosh

1/(tan2x+sin2x)

there's your problem....

Reply 2

habosh

OP

13

elpaw

there's your problem....

How...doesn't cot2x = 1/tan2x
and cosec 2x=1/sin2x???

Reply 3

elpaw

15

it does.... but that is not what you have written....

Reply 4

habosh

OP

13

elpaw

it does.... but that is not what you have written....

Oh I get it...stupid me :tongue:

thanx

Reply 5

username9816

14

habosh

Prove that
Cot2x+cosec2X=cotX

LHS = Cot2x + cosec2X =
1/tan2x + 1/sin2x =
cos2x/sin2x + 1/sin2x =
cos2x/2sinxcosx + 1/2sinxcosx =
(cos2x + 1)/2sinxcosx =
(2cos^2x - 1 + 1)/2sinxcosx =
2cos^2x/2sinxcosx =
cos^2x/sinxcosx =
cosx/sinx =
1/tanx =
cotx =
RHS

QED

Reply 6

habosh

OP

13

Invisible

LHS = Cot2x + cosec2X =
1/tan2x + 1/sin2x =
cos2x/sin2x + 1/sin2x =
cos2x/2sinxcosx + 1/2sinxcosx =
(cos2x + 1)/2sinxcosx =
(2cos^2x - 1 + 1)/2sinxcosx =
2cos^2x/2sinxcosx =
cos^2x/sinxcosx =
cosx/sinx =
1/tanx =
cotx =
RHS

QED

I solved it in another way but I got the answer right :biggrin:

...I don't get why 1/tan2x = cos2x/sin2x....I've never saw that in my heinemann book
thanx anyway :tongue:

Reply 7

elpaw

15

habosh

I solved it in another way but I got the answer right :biggrin:

...I don't get why 1/tan2x = cos2x/sin2x....I've never saw that in my heinemann book
thanx anyway :tongue:

tan2x = sin2x/cos2x => 1/tan2x = cos2x/sin2x

Reply 8

habosh

OP

13

Invisible

tan2x = sin2x/cos2x

Agree? Ok.

oh that was my problem..I get now
*starts memorising the formula* :smile:

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