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Author Topic: [CIE] All Pure Mathematics (P1, P2 and P3) doubts here !  (Read 81809 times)

Offline Arthur Bon Zavi

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #30 on: November 08, 2010, 05:21:02 am »
You are right DK.

\frac{1}{cos x} + \{tan x}\equiv\frac{cos x}{1 - sin x}

\frac{1}{cos x} + \frac{sin x}{cos x}

\frac{cos x + (sin x) (cos x)}{cos^2 x}

\frac{cos x (1 + sin x)}{1 - sin^2 x}  :'(

\frac{cos x (1 + sin x)}{(1 - sin x) (1 + sin x)}

(1 + sin x) and (1 + sin x) cancel out.

Therefore :

\frac{cos x}{1 - sin x}

Proved the Identity :

\frac {1}{cos x} + {tan x}\equiv\frac{cos x}{1 - sin x}
« Last Edit: November 08, 2010, 05:36:44 am by Deadly_king »
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Offline Arthur Bon Zavi

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #31 on: November 14, 2010, 09:29:45 am »
Prove the Identities :

(i) \frac{1}{Tan x} + tan x\equiv\frac{1}{Sin x Cos x}


(ii) \frac{1 - 2 sin^2 x}{Cos x + Sin x}\equiv Cos x - Sin x

After trying much, I didn't get !
« Last Edit: November 14, 2010, 10:50:38 am by Ancestor »
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Alpha

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #32 on: November 14, 2010, 10:57:37 am »
1) (1+tan^2 x)/ tan x = sec^2 x/ tan x

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

Check the second part too... I think there should be a double angle somewhere.
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Offline Arthur Bon Zavi

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #33 on: November 14, 2010, 11:14:27 am »
Quote from: ~Alpha on November 14, 2010, 10:57:37 am
1) (1+tan^2 x)/ tan x = sec^2 x/ tan x

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

Check the second part too... I think there should be a double angle somewhere.

You took L.H.S, I don't think so !  :-\
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Alpha

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #34 on: November 14, 2010, 11:17:37 am »
1/ tan x + tan x = 1+tan^2 x)/ tan x
                      = sec^2 x/ tan x

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

-_-
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Offline Arthur Bon Zavi

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #35 on: November 14, 2010, 11:22:56 am »
Quote from: ~Alpha on November 14, 2010, 11:17:37 am
1/ tan x + tan x = 1+tan^2 x)/ tan x
                      = sec^2 x/ tan x

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

-_-

Your head must be aching. Sorry, for making you run behind this. :-X

Thank You. :D
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Alpha

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #36 on: November 14, 2010, 11:30:26 am »
Quote from: Ancestor on November 14, 2010, 11:22:56 am
Your head must be aching. Sorry, for making you run behind this. :-X

Thank You. :D

You're welcome. :)

Yes, actually, my head IS aching, ain't cause of your Maths question though.  ;)
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Offline $H00t!N& $t@r

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #37 on: November 16, 2010, 03:17:48 pm »
Find the value of k for which there is no term in x^2 in the expansion of (1+kx)(2-x)^6          :)
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Offline Arthur Bon Zavi

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #38 on: November 16, 2010, 03:22:28 pm »
Quote from: $H00t!N& $t@r on November 16, 2010, 03:17:48 pm
Find the value of k for which there is no term in x^2 in the expansion of (1+kx)(2-x)^6          :)

Post whole Question, from the first ! If you can, tell us the page number of the P1 book and question number.
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elemis

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #39 on: November 16, 2010, 03:29:41 pm »
Quote from: $H00t!N& $t@r on November 16, 2010, 03:17:48 pm
Find the value of k for which there is no term in x^2 in the expansion of (1+kx)(2-x)^6          :)

Doing it.
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elemis

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #40 on: November 16, 2010, 03:37:44 pm »
Expanding (2-x)6 up to the term containing x2 gives :

64 - 192x +240x2

Multiply this by (1+kx) giving :

64 - 192x + 240x2 +64kx -192kx2 +240kx3

Since x2 should not be present the sum of the co-efficients of the x2 terms must add up to zero.

Therefore, 240 -192k = 0

k = 1.25
« Last Edit: November 16, 2010, 03:42:20 pm by Ari Ben Canaan »
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Offline $H00t!N& $t@r

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #41 on: November 16, 2010, 03:46:52 pm »
thanks alot Ari  ;D
+rep
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elemis

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #42 on: November 16, 2010, 03:48:39 pm »
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Offline $H00t!N& $t@r

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #43 on: November 17, 2010, 10:18:33 am »
how do you find the range of a fuction?  ??? i read the book but i dont understand it  :-\

can someone help me with m/j 05 q7 (i) and (iii)  please explain the answer in (iii) Thanks
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"If A equals success, then the formula is:
A=X+Y+Z, X is work. Y is play. Z is keep your mouth shut"- Albert Einstein

elemis

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #44 on: November 17, 2010, 10:24:10 am »
Quote from: $H00t!N& $t@r on November 17, 2010, 10:18:33 am
how do you find the range of a fuction?  ??? i read the book but i dont understand it  :-\

can someone help me with m/j 05 q7 (i) and (iii)  please explain the answer in (iii) Thanks

Sorry, I cant do this, havent done it yet.
« Last Edit: November 17, 2010, 10:33:47 am by Ari Ben Canaan »
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