...ln|sec(t)| -2*ln|sin(t)| + C w = C*sec(t)/sin^2(t) = v' v = ∫

**Csec**(t)/sin^2(t)) dt = ∫Ccos(t)/(1 - sin^2(t))*(sin^2(t)) dt Let z...1 Answers · Science & Mathematics · 30/09/2013

...d/dx[ycos(x)] = cos(x) ycos(x) = sin(x) + C y = tan(x) +

**Csec**(x) -------------- 2.) 1/√(x^2 - 3) (dy/dx) = xcos^2(y) sec^2(y) dy = x...1 Answers · Science & Mathematics · 25/02/2013

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

...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

...xsin(x) - ∫ sin(x)dx ycos(x) = xsin(x) + cos(x) + C y = xtan(x) +

**Csec**(x) + 1 answer//3 Answers · Science & Mathematics · 20/08/2011

y' - (tan x)y = 1 y ' cos x - (sin x) y = cos x y ' cos x +(cos x)' y = cos x (y cos x) ' = cos x y cos x = sin x +C y = (sin x +C)/cos x y = tan(x) +

**Csec**(x)1 Answers · Science & Mathematics · 09/03/2010

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

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

...hence we can find perpendicular = 3 hence tan A = perpendicular/ base = -3/2

**Csec**is 1/ sinA and Sin A = p/h = 3/5sp**Csec**...4 Answers · Science & Mathematics · 30/08/2009

Just keep taking the derivatives until it isn't 0/0, you're allowed to do that. With the x^3 on the bottom you'll need to apply L'Hopital's rule 3 times to get this not to be zero (x^3 -> 3x^2 -> 6x -> 6.)

1 Answers · Science & Mathematics · 22/04/2009

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