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related to: CSEC

1. ### what is the difference between the term csec ,sec,and arcsin arccos math?

Cosec and sec are the reciprocals of SIN and COS. Arc cos(X) is the angle whose cosine is X Arc sin(X) is the angle whose sin is X

5 Answers · Science & Mathematics · 13/02/2009

2. ### I want a site to fine my cxc results ?

In Trinidad, CSEC results will be released to schools either on Thursday...

5 Answers · Science & Mathematics · 25/08/2008

3. ### Use implicit differentiation to find dy/dx?

dx+ csec ^2(xy).( xdy + ydx)=0 dy(x csec ^2(xy)=(-1-ycsec^2(xy))dx dy/dx=(-1-ycsec^2(xy))/(x csec ^2(xy)

4 Answers · Science & Mathematics · 28/10/2012

4. ### verify that cscx / (cotx + tanx ) = cos x?

cscx / (cotx + tanx ) = csec x /( 1/tan x + tan x) = csc x / ((1+ tan ^2 x)/tan x) = csc x tan x / (sec ^2 x) = (1/sin x) (sin x/cos x) cos ^2 x = cos ^2 x/ cos x = cos x proved by math kp

2 Answers · Science & Mathematics · 02/06/2008

5. ### How to write maths SBA?

math SBA? are you doing CAPE or CSEC mathematics? if so, then the SBA component for CAPE is 3 module tests (i think) CSEC Math doesnt have any SBAs

1 Answers · Science & Mathematics · 29/10/2007

6. ### Is there and easier way to do trig identities?

Yes Observe sin(x) cos(x) tan(x) cot(x) sec(x) csec (x) 1/sin(x)= csec (x) 1/cos(x)=sec(x) 1/tan(x) =cot(x) also sin(x)/cos(x)=tan...

4 Answers · Science & Mathematics · 24/01/2007

7. ### Is it possible to make a TI-84+ program where it wil appear in the math menu and not run as a program?

No. a TI-Basic program doesn't have any where near the capability of doing what you want. You would have to program it in Z80 Assembly, which is by far too tedious. If you have a cable that connects the TI-84 Plus to...

1 Answers · Science & Mathematics · 25/01/2010

...2(x)]/cos(x)(x) - Dsin(x)[tan^2(x)] = Acos(x) - Bsin(x) + Csec (x) - Ccos(x) - Dsin(x)*[sec^2(x) - 1] = Acos(x) - Bsin...

1 Answers · Science & Mathematics · 05/05/2012

9. ### Prove that tanx + cotx = (secx)(cscx).?

left part tan x = sinx / cosx cot x = cosx / sinx sin²x + cos²x = 1 ((sinx)² + (cosx)²) / (cosx*sinx) = 1 / (cosx* sinx) right part sec = 1 / cosx csec = 1 / sinx 1 / (sinx*cosx) so left = right

4 Answers · Science & Mathematics · 09/01/2010

10. ### simplify: (( tan x) / ( csc x + cot x))?

becuase csc^2 x -c cot^2 x = 1 multiply both num and den by csc x - cot x = tan x( csec x - cot x) = tan x csc x - 1 = sinx / cos x * 1/ sin x -1 = sec x -1 so ans is a

2 Answers · Science & Mathematics · 24/03/2007