IGCSE/GCSE/O & A Level/IB/University Student Forum

Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: nipuna92 on May 11, 2010, 03:34:36 am

Title: Proving an identity (P1) trignometry
Post by: nipuna92 on May 11, 2010, 03:34:36 am

Not sure if the question is correct
anyway here it is

http://s941.photobucket.com/albums/ad251/flameir/Q2/?action=view&current=10052010048.jpg

Please try to include all ur steps in doing this question

Title: Re: Proving an identity (P1) trignometry
Post by: Monica on May 11, 2010, 04:57:03 am

Quote from: nipuna92 on May 11, 2010, 03:34:36 am

Not sure if the question is correct
anyway here it is

http://s941.photobucket.com/albums/ad251/flameir/Q2/?action=view&current=10052010048.jpg

Please try to include all ur steps in doing this question

I will solve it and post in 10 mins.

Title: Re: Proving an identity (P1) trignometry
Post by: nid404 on May 11, 2010, 05:23:28 am

Well...I'm done so ill post as well :)

Anyway this would have to be broken into steps because it's quite long compared to the ones we get now....seems like an old paper

Let's deal with the numerator first

sin²x -cos²x

sin²x can be written as cos²x X tan ²x

that is because tanx=sinx/cosx

now the numerator looks like this

cos²x X tan ²x -  cos²x

taking  cos²x common you get

 cos²x ( tan ²x - 1)

Now leave it at this. Move to the denominator

1+ 2sinxcosx
we know that sinx=cosxtanx

let's replace sinx we get

1+ 2 cosxtanx X cosx

1+ 2cos²xtanx

we also know
sin²x + cos²x =1

let's replace 1 with this

sin²x + cos²x + cos²xtanx

Now we still need to replace sin²x again by cos²x tan ²x

Doing so you get

cos²x tan ²x+ cos²x + 2cos²xtanx

Now take cos²x  common

it reduces to
cos²x (1+ tan²x + 2tanx)

Now bring in the numerator again

cos²x ( tan ²x - 1)

divide the two
cos²x ( tan ²x -1)
----------------------------------------
cos²x (1+ tan²x + 2tanx)

cos²x gets cancelled

remaining bring it down again

 ( tan ²x - 1)------->  (tanx+1)(tanx-1)   1)
------------------------
(1+ tan²x + 2tanx)------>(tanx+1)²  2)

Divide 1 by 2

(tanx+1)(tanx-1)
-----------------
(tanx+1) (tanx+1)

(tanx+1)(tanx-1)
------------------------
(tanx+1) (tanx+1)

You have your answer
(tanx-1)
--------
(tanx+1)

Title: Re: Proving an identity (P1) trignometry
Post by: Monica on May 11, 2010, 05:34:14 am

Quote from: nid404 on May 11, 2010, 05:23:28 am

Well...I'm done so ill post as well :)

Anyway this would have to be broken into steps because it's quite long compared to the ones we get now....seems like an old paper

Let's deal with the numerator first

sin²x -cos²x

sin²x can be written as cos²x X tan ²x

that is because tanx=sinx/cosx

now the numerator looks like this

cos²x X tan ²x -  cos²x

taking  cos²x common you get

 cos²x ( tan ²x - 1)

Now leave it at this. Move to the denominator

1+ 2sinxcosx
we know that sinx=cosxtanx

let's replace sinx we get

1+ 2 cosxtanx X cosx

1+ 2cos²xtanx

we also know
sin²x + cos²x =1

let's replace 1 with this

sin²x + cos²x + cos²xtanx

Now we still need to replace sin²x again by cos²x tan ²x

Doing so you get

cos²x tan ²x+ cos²x + 2cos²xtanx

Now take cos²x  common

it reduces to
cos²x (1+ tan²x + 2tanx)

Now bring in the numerator again

cos²x ( tan ²x - 1)

divide the two
cos²x ( tan ²x -1)
----------------------------------------
cos²x (1+ tan²x + 2tanx)

cos²x gets cancelled

remaining bring it down again

 ( tan ²x - 1)------->  (tanx+1)(tanx-1)   1)
------------------------
(1+ tan²x + 2tanx)------>(tanx+1)²  2)

Divide 1 by 2

(tanx+1)(tanx-1)
-----------------
(tanx+1) (tanx+1)

(tanx+1)(tanx-1)
------------------------
(tanx+1) (tanx+1)

You have your answer
(tanx-1)
--------
(tanx+1)

Thank You! I couldn't really get it..I have a long answer  which I still didn't get into conclusion.

Thanks A LOT Nid! InshAllah you will get an A in this subject. =]

Title: Re: Proving an identity (P1) trignometry
Post by: nid404 on May 11, 2010, 05:35:26 am

No problem dear. Hope you got it now :)

Ahh...thank you for that. I really hope I do :)

Wish you luck!! You'll do great