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[tan(t) +

**sec**(t) - 1] / [tan(t) -**sec**(t) + 1] = tan(t) +**sec**(t...left hand side (LHS). LHS = [tan(t) +**sec**(t) - 1] / [tan(t) -**sec**(t) + 1] Change everything...t) By definition, LHS = tan(t) +**sec**(t) = RHS1 Answers · Science & Mathematics · 02/03/2007

**sec**(x) - sin(x) = [tan^2(x) + cos^2(x)] / [**sec**(x) + sin(x)] LS:**sec**(x) - sin(x) = [**sec**(x) - sin...x) = 1 dividing through by cos^2(x) tan^2(x) + 1 =**sec**^2(x) tan^2(x) =**sec**^2(x) - 1 resubstituting tan^2(x): =**sec**...2 Answers · Science & Mathematics · 30/11/2009

x * tan(x) * dx / (

**sec**(x) + tan(x)) => x * tan(x) * (**sec**(x) - tan(x)) * dx / (**sec**(x)^2 - tan(x)^2) => x * (tan(x) ***sec**(x) - tan(x)^2) * dx / 1 => x * (**sec**(x) * tan(x) -**sec**(x)^2 + 1) * dx u = x du = dx dv =**sec**(x) * tan(x) * dx -**sec**(x...3 Answers · Science & Mathematics · 29/04/2013

∫

**sec**[(5/2)x] tan[(5/2)x] dx = (I assume x is in the numerator) let [(5/2)x] = u → x = (2/5)u → dx = (2/5) du then, substituting, you get: ∫**sec**(5/2x) tan(5/2x) dx = ∫**sec**u tan u (2/5) du = take out the constant: (2/5) ∫**sec**...3 Answers · Science & Mathematics · 14/08/2008

y*

**sec**(x) = x*tan(y) Differentiating both sides with respect to x, y * d/dy(**sec**(x)) +**sec**(x) * dy/dy = x * d/dy(tany) + tan(y) * dx/dy Or y * d/dy(**sec**(x...2 Answers · Science & Mathematics · 18/07/2009

The

**SEC**has only lost 1 BCS game since 2001. By contrast, Choke...also blistered the ACC champion by 41 points. The**SEC**'s bowl record is more impressive when considering they play few ...18 Answers · Sports · 03/01/2008

1+

**sec**(x)=tan(x) 1+**sec**(x)=sin(x)/cos(x) Crossing multiplication. cos(x)(1+**sec**(x))=sin(x) (cos(x)+1)^2=(sin(x))^2 cos^2(x)+2cos(x)+1=sin^2(x) ..... ... ...11 Answers · Science & Mathematics · 02/08/2013

First, a few identities: [a]

**sec**x = 1 / cos x [b] csc x = 1 / sin x [c] cos x + n360 = cos x...90. 5) Apply [i], [j] or [k] if possible. So...**sec**-150 = 1/cos -150 [a] = 1/cos 150 [e] = -(1/cos 30) [g] = -(1...1 Answers · Science & Mathematics · 03/12/2007

∫

**sec**(√x) / √x dx let √x =t (1(/2√x))*dx=dt substitute ∫**sec**(√x) / √x dx=∫ 2**sec**t dt =2ln|**sec**t+tant|+C t=√x hence ∫**sec**(√x) / √x dx...2 Answers · Science & Mathematics · 22/06/2013

**sec**^2(18) - tan^2(18) First off, remember that there is an identity that relates**sec**^2(x) with tan^2(x); it goes**sec**^2(x) = tan^2(x) + 1. Therefore...6 Answers · Science & Mathematics · 13/03/2007

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