sec^4y-tan^4y/sec^2y+tan^2y=1?
929231
(a^2-b^2)=(a-b)(a+b)\nHence,\nsec^4y-tan^4y = (sec^2y-tan^2y)(sec^2y+tan^2y)\n\nSubstitute that back into the initial equation...\n\n(sec^2y-tan^2y)(sec^2y+tan^2y)\n------------------------------------------\n           (sec^2y+tan^2y)\n\nand cancel common terms\n\n= (sec^2y-tan^2y) \n= 1 (a standard trig identity)\n\nhope that solves it.
482644
Let us denote the integral sign by I.\nWe will use substitution method.\nLet x = 2*tan(t)\nThen dx = 2*sec^2(t)dt\nThus, I[1/(x^2+4)^2]dx\n= I[1/{4*tan^2(t)+4}^2]*2*sec^2(t)dt\n= (1/8)*I[1/{tan^2(t)+1}^2]*sec^2(t)dt\n= (1/8)*I[1/{sec^2(t)}^2]*sec^2(t)dt  [since tan^2(t) + 1 =sec^2(t)]\n= (1/8)*I[1/{sec^2(t)}]dt\n= (1/8)*I[cos^2(t)]dt\n= (1/16)*I[2*cos^2(t)]dt\n= (1/16)*I[1+cos(2t)]dt [since 2*cos^2(t) = 1 + cos(2t)]\n= (1/16)*I[dt] + (1/16)*I[cos(2t)]dt\n= (1/16)*t + (1/16)*[{sin(2t)}/2] + C where C = arbitrary constant of integration\n= t/16 + {sin(2t)}/32 + C\n\nNow sin(2t) \n= {2*tan(t)}/{1+tan^2(t)}\n= 4x/(4+x^2) [since x = 2*tan(t)]\n\nAlso t = tan-1(x/2) [tan-1(x/2) means tan inverse (x/2)]\n\nHence the given integral equals to\n{tan-1(x/2)}/16 + x/{8(4+x^2)} + C
216863
((4 * 4) / 4) - 4 + 4 = 4
971390
{3XY^2M/1}-{2M/[X^-1Y^-1]^2\n+{[4X^2Y^2]/M^-1}-\n(3X^(3Y)^(3M)^-1/(XYM)^-2
470919
I think you need to find A, if it is so then:\ntan A + sec A=2     square both sides\ntan^2 A + 2 tan A sec A + sec^2 A=4  since, sec^2 A= 1+tan^2 A  then:\n2 tan^2 A + 2 tan A sec A=3      since tan A=sin A/ cos A & sec A=1/cos A   then:\n2 sin^2 A/ cos^2 A + 2 sin A/ cos^2 A=3\n sin^2 A/ cos^2 A + sin A/ cos^2 A=3/2\n sin^2 A + sin A= 3/2* cos^2 A = 3/2* (1-sin^2 A)    since sin^2 A + cos^2 A =1\n5/2* sin^2 A + sin A -3/2= 0\n5 sin^2 A + 2 sin A – 3 = 0\n(5 sin A -3) (sin A +1)= 0 then,\neither sin A= 3/5 → A=36.87o \nor sin A=-1 → A=-90o (to be neglected)\ncheck: tan 36.87o=0.75 & cos 36.87o = 0.8 → sec 36.87o =1.25 then,\ntan 36.87o  + sec 36.87o =2  Q.E.D
714113
secant x              = 1 / cos x\nderivative             = [cos x (0) - 1 (-sin x)] / cos2 x     (I mean cos square)\n                           =  sinx / cos2 x\n                           =  tan x / cos x                (sin x / cos x = tan x)\n                           = tan x sec x                   (1 / cos x = sec x)\n\nDone! Hope this can help.
744903
1)\n\ntan^2(x) = sec^2(x) - 1\n\nSo the integral of sec^2(x) + 4 = tan(x) + 4x\n\n2)\n\ncsc(x) = 1/sin(x)\n\nthe integral is then of 1+sin(x)/cos^2(x).\nlet u = cos(x) then du = -sinx dx.\nso the integral is x+sec(x).\n\n3)\n\nlet u = cos(2x+1) then du = -(1/2)sin(2x+1) dx.\n\nso the integral is of -(1/2)/u^2 = 2/u = (1/2)/cos(2x+1) = sec(2x+1)/2\n\nSteve
933478
Using the Wolfram Integrator:\nhttp://integrals.wolfram.com/index.jsp\n\nOne obtains\n1/10 sec(2x)tan^4(2x) + 2/15 sec(2x)tan^2(2x) + 4/15 tan(2x)\n\nUsing the other poster's logic - you get (u^2 + 1)^3
591905
1+2(tanx)^2+(tanx)^4\n= (1+(tanx)^2)^2\n= ((secx)^2)^2\n= (secx)^4\n= (cosx)^-4
697906
let y = arc tan (x)\n\ntan y = x\ndifferentiate\n\nsec^2 y dy/dx = 1\n\nor(1+tan ^2y) dy/dx = 1\n\ndy/dx = 1/(1+tan^2 y) = 1/(1+x^2)
235956
tan(nPi/2) when n=7\n\nor tan(nPi/4) when n=2\n\nor if you need one expression,\n\n1/[Tan(nPi/2) Tan(nPi/4)].\n\nRemember, tan = sin/cos...
1086900
Yes it is true for x=200.62 degrees.\n\nCarefuly schetch tan(x) and sec(x) [sec(x)=1/cos(x)] and you will see the answer or let your fingers do the calculator walk :).\n\n\nWas that helpfull?
859744
(x + 4y)(x - 4y)\nI answered 2 of your questions. Do the rest of the questions yourself.
982428
LHS\n= (sin^2B-tan^2B)/(1-sec^2B)\n= (sin^2B-tan^2B)/(-tan^2B)\n= (sin^2B/-tan^2B) + (-tan^2B/-tan^2B)\n[splitting the numerator up]\n= (-cos^2B) + 1\n[cosx= sinx/tanx]\n= 1 - cos^2B\n= sin^2B\n= RHS
710286
the antiderivative of 4 / (t^2 + 1)  IS \n4 arctan(t) + C`
714467
(2 tan x ) / (3 - 2 tan squared x) = 1\n\nMultiply both sides by  (3 - 2 tan squared x)  and get\n\n2 tan x = 3 - 2 tan squared x\n\nLet u= tan x and the equation becomes\n\n2 u= 3 - 2 u^2\n2u^2 +2u -3 = 0\nu^2 + u -1.5=0\n\ncompleting the square you get\n(u  + 0.5)^2 -1.5 - 0.25 = 0\n(u  + 0.5)^2 -1.75= 0\n(u  + 0.5)^2 =1.75\nu+0.5 = square root (1.75)\nu +0.5=1.32287566\nu=1.32287566-0.5\nu=0.82287566\n\ntan x = 0.82287566\n\nx=arctan ( 0.82287566)\nx=0.688534708 radians \n\nas the tan shape repeats \n\nx = 0.688534708 radians AND (0.688534708 +pi) radians \nx = 0.688534708  and 3.83012736 radians
592160
i think the answer wil be (tanX)/(tanY)\n\nSIN(X+Y)=sin x cos y+cos x sin y\n\nsin(x-y)=sin x cos y-cos x sin y\n\n use cos x cos y to divide the numerator and denominator... there is the answer...