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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

In Trinidad,

**CSEC**results will be released to schools either on Thursday...5 Answers · Science & Mathematics · 25/08/2008

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

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 kp2 Answers · Science & Mathematics · 02/06/2008

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 SBAs1 Answers · Science & Mathematics · 29/10/2007

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

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

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 = right4 Answers · Science & Mathematics · 09/01/2010

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 a2 Answers · Science & Mathematics · 24/03/2007

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