differenciate half of tanx sinx?
408737
Need to use the product rule here:\n\nd/dx(1/2*tanxsinx)\n\n= [d/dx(1/2tanx)]sinx + 1/2tanx*d/dx(sinx)\n= (1/2sec^2x)sinx + 1/2tanx*cosx\n= sinx/2cos^2x + sinx/2cos^2x\n= tanx/cosx
837557
(secx - tanx)^2 = (1/cosx - sinx/cosx)^2 = ((1-sinx)/cosx)^2\n\n(1-sinx)/(1+sinx) \n= [(1-sinx)(1-sinx)/(1+sinx)(1-sinx)]\n\n= [(1-sinx)^2/(1-(sinx)^2)] \n= [(1-sinx)^2/(cosx)^2] \n= ((1-sinx)/cosx)^2
575459
Sorry but I don't quite understand your qn.\nCan you make it clearer?\n\n(to ozert:\ncos A and sin A could be negative values if tan A is in Quadrant 3)
676590
Sorry, I can't see the photo...but:\n\nSOH - sin = opp/hyp\nCAH - cos = adj/hyp\nTOA - tan = opp/adj
932836
Are you sure this is the correct identity?\n\n 1+ tanx.tan2x = 1 + (sinx/cosx)(sin2x/cos2x)\n                        = 1 + (sinx/cosx)(2sinxcosx/cos2x)\n                        = 1 + 2sinx.sinx/cos2x\n\n cos2x = 2cosx.cosx - 1  = 2cos^2x - 1        (Note:  ^2 = square)\n\n then..\n      1+tanx.tan2x = 1 + [2sin^2x/(2cos^2x - 1)]\n                            = (2cos^2x + 2sin^x -1) / (2cos^2x - 1)\n                            = [2(cos^2x + sin^2x) -1] / (2cos^2x - 1)\n                            = 1/cos2x    (because 2cos^2x - 1 = cos2x)\n\n     Then 1/1+tanxtan2x = cos2x
860568
∫tan^(1/3)x dx\nI can predict that this'll involve the use of partial fractions.\n\nLet tanx=u^3\n\n=>(sec x)^2 dx=3u^2 du\n\nnow, (sec x)^2=1+ (tan x)^2=1+u^6\n\n=>dx=(3u^2)/1+ u^6\n\ntherefore, ∫tan^(1/3)x dx=3∫[u X u^2/(1+u^6)]du\n\n=>3∫[u^3/(1+u^6)]du\nlet u^2=t\n=>(2u) du=dt\n\n=>3∫[u^3/(1+u^6)]du=(3/2)∫[t/(1+t^3)]dt\n\nNow, partial fractions will be used\nwe know that, t/(1+t^3)=t/(1+t)(1+t^2-t)...........since a^3 + b^3=(a+b)(a^2 + b^2 + ab)\n\nlet, t/(1+t)(1+t^2-t)=A/(1+t) + (Bt + C)/(1+t^2-t)\n\n=>t/(1+t)(1+t^2-t)=[A(1+t^2 -t) + (Bt + C)(1+t)]/(1+t)(1+t^2 -t)................taking LCM\n\nequating coefficients of u, u^2, constants on both sides\n=>1=-A+B+C...................equating coefficients of u\n     0= A+B .......................equating coefficients of u^2\n     0= A+C .......................equating coefficients of constant\n=>A=-1/3, B=1/3, C=1/3\n\n=> (3/2)∫[t/(1+t^3)]dt=\n(3/2)∫[(1/3)/(1+t)]dt + (3/2)∫[(1/3)(1+t)/(1+t^2-t)]dt\n\n\n=>(-1/2)log|1+t| +  (1/2∫[(1+t)/(1+t^2-t)]dt\n\n=>(-1/2)log|1+t| +  (1/4)∫[(2+2t)/(1+t^2-t)]dt \n\nwe know, d/dt(1+ t^2 - t)=2t -1\n=>(-1/2)log|1+t| +  (1/4∫[(2+2t+1-1)/(1+t^2-t)]dt \n\n=>(-1/2)log|1+t| +  (1/4)log|1+t^2 -t| + (3/4)∫1/(1+t^2-t)dt\n\nwe know, 1+t^2-t=(t-1/2)^2 +3/4\n\n=>(-1/2)log|1+t| +  (1/4)log|1+t^2 -t| + (3/4)∫1/[(t-1/2)^2 + 3/4]dt\n\n=>(-1/2)log|1+t| +  (1/4)log|1+t^2 -t| + \n(3/4)1/(2/sqrt(3) arctan[2(t-1/2)/sqrt(3)\n\n\nsubstituting original values t=u^2\n\n=>(-1/2)log|1+u^2| +  (1/4)log|1+u^4 -u^2| + \n(3/4)1/(2/sqrt(3) arctan[2(u^2-1/2)/sqrt(3)\n\nsubstituting original values u=(tanx)^1/3\n\n=>(-1/2)log|1+(tanx)^2/3| +  (1/4)log|1+(tanx)^4/3 -(tanx)^2/3| + \n(3/4)1/(2/sqrt(3) arctan[2((tanx)^2/3 - 1/2)/sqrt(3) + C\n\nwhere arctan(x) means tan inverse x\nresults used= \n1)∫1/(1+x)dx= log|1+x| + c\n2)∫1/(a^2+x^2)dx=\n(1/a)arctan(x/a) + C...........'a' is a constant\n\n\nP.S. Extremely sorry for the length of the solution but it is correct and each and every explanation is given.
1117571
less than half...........
1118472
Move the decimal two places to the right.\n83.4%
1105851
8....it's half.\n\nEach reduce down to 1/2.
702823
want the proof? Here it is ( i am assuming you know the dervatives of sinx and cosx)\n\ny = tanx\ny = sinx/cosx\ny' = [cosx(cosx) - sinx(-sinx)]/(cosx)^2 --> quotient rule\ny'= [(cosx)^2 + (sinx)^2]/(cosx)^2\ny' = 1/(cosx)^2\ny' = (secx)^2\n\nvoila
810960
Easy. 40 divided by 2 is 20. Take half of 40.  OR 40x0.5
443025
(tanA + tanB + tanC - tanAtanBtanC) / (1 - tabAtabB - tabBtanC - tanCtanA)\ngets nasty fast, so remember these rules\ntanX = sinX / cosX\nsin(X+Y) = sinXcosY + cosXsinY\ncos(X+Y) = cosXcosY - sinXsinY\n\n(tanA + tanB + tanC - tanAtanBtanC) / (1 - tabAtabB - tabBtanC - tanCtanA)\n=[cosAcosBcosC * (tanA + tanB + tanC - tanAtanBtanC)] / [(1 - tabAtabB - tabBtanC - tanCtanA) * cosAcosBcosC]\n=[sinAcosBcosC + sinBcosAcosC + sinCcosAcosB - sinAsinBsinC] / [cosAcosBcosC - sinAsinBcosC - sinBsinCcosA - sinCsinAcosB]\n=[(sinAcosC + cosAsinC)cosB + sinB(cosAcosC - sinAsinC)] / \n[cosC(cosAcosB - sinAsinB) - sinC(sinAcosB + cosAsinB)]\n=[sin(A+C)cosB + sinBcos(A+C)] / [cosCcos(A+B) - sinCsin(A+B)]\n=sin(A+B+C) / cos(A+B+C)\n=tan(A+B+C)
925426
one half and seven eightths
1021739
Since tanx = sinx/cosx,\n\ntan(A-B) = sin(A-B)/cos(A-B)\n\nNow, write out the expression for each term on the right side of this equation.\n\nWhen you finish that, divide each term by cosAcosB and you will get the result you are seeking.
605811
As Lauren said, it's 1/x.\n\nMemorize this. It will be very useful.
201111
You need to measure the distance between the closely spaced lines (J values).  Trans and cis J values will be different
952344
(1+tanx)/(1+cotx)\n=(1+sinx/cosx)/(1+cosx/sinx)\n=[(cosx+sinx)/cosx]/(sinx+cosx)/sinx]\n=(cosx+sinx)(sinx)/(cosx+sinx)cosx)\n=sinx/cosx\n=tanx\nhence proved