prove that tan^-1 1/3+ tan^-1 1/5+ tan^-1 1/7+ tan-1 1/8 = pi/4?
540401
(3 + i)(5 + i)(7 + i)(8 + i) = 650 + 650i\n\nwhere i is the square root of -1.
766844
1/tanx - 1/x=\n(x-tanx)/(xtanx)=0/0\nso use l'hopital's rule\n(1-sec^2(x))/(tanx+xsec^2(x))=\n(1-1)/(0+0)= 0/0\nso use l'hopital again\n(-2secxsecxtanx)/(sec^2(x)+\nsec^2(x)+x(2secxsecxtanx))=\n0/(2+0)=0
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
662965
integral of (tan(x))^(1/3) ... hmmm ...\n\nSubstitute u = tan(x)\nso that du = sec^2(x) dx = (1 + tan^2(x)) dx = (1 + u^2) dx\nand so dx = du / (1 + u^2)\n\nThe substitution therefore yields the\nintegral of (u^(1/3) / (1 + u^2)) du\n = integral du u^(1/3)(1 - u^2 + u^4 - u^6 + u^8 - ...)\n = integral du (u^(1/3) - u^(7/3) + u^(13/3) - u^(19/3) + ...)\n\n = (3/4) u^(4/3) - (3/10) u^(10/3) + (3/16) u^(16/3) - ... + C\n\n = 3 u^(1/3) ( u/4 - (u^3)/10 + (u^5)/16 - ... )\n\nwhich can be written as\n\n3 u^(1/3) sum{ n = 0 to infinity of (-1)^n u^(2n+1) / (6n+4) }\n\nthen do the "+ C" thing and replace u by tan(x)\n\nI would estimate -- an educated guess -- that the radius of convergence of u = tan(x) is 1.
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
859324
1) 8 - 3 4/5 = 5 - 4/5 = 4 + 1 - 4/5 = 3 1/5\n2) 2 1/6 - 1 2/3 = 1 + 1/6 - 2/3 = 1 1/6 - 4/6 = = 1 - 3/6 = 1 - 1/2 = 1/2\n3) 6 1/10 - 4 3/10 = 2 + 1/10 - 3/10 = 2 - 2/10 = 1 + 1 - 1/5 = 1 4/5\n4) 3 1/8 - 2 1/4 = 1 + 1/8 - 1/4 = 1 + 1/8 - 2/8 = 1 - 1/8 = 7/8\n5) 4/4 - 4 7/8= 1 - 4 7/8 = -3 - 7/8 = -3 7/8
167162
I do agree with Anniemc46.\nIf the denominator is 1+(tanhx)^2 then it is easy to do.\n\nI have done and uploaded to my web site.\nPlease check the url I provided in the sourse.\n\nThanking U,\nThuvalpakshi.
137846
2 1/2 x 1 1/3 =5/2 x 4/3=10/3 = 3 1/3
817617
tan(A-B)+tan(B-C)+tan(C-A)\n= (tanA - tanB)/(1 + tanAtanB) + (tanB - tanC)/(1 + tanBtanC) + (tanC - tanA)/(1 + tanAtanC)\n= [(tanA - tanB)(1 + tanBtanC)(1 + tanAtanC) + (tanB - tanC)(1 + tanAtanC)(1 + tanAtanB) + (tanC - tanA)(1 + tanAtanB)(1 + tanBtanC)]/[(1 + tanAtanB)(1 + tanBtanC)(1 + tanAtanC)]\n= 0\nie (tanA - tanB)(1 + tanBtanC)(1 + tanAtanC) + (tanB - tanC)(1 + tanAtanC)(1 + tanAtanB) + (tanC - tanA)(1 + tanAtanB)(1 + tanBtanC) = 0\nOn expanding\n(tanA - tanB)(1 + tanAtanC + tanBtanC + tanAtanBtan²C) \n+ (tanB - tanC)(1 + tanAtanC + tanAtanB + tan²AtanBtanC)\n+ (tanC - tanA)(1 + tanAtanB + tanBtanC + tanAtan²BtanC)\n\n= tanA - tanB + tan²AtanC - tanAtanBtanC + tanAtanBtanC - tan²BtanC + tan²AtanBtan²C - tanAtan²Btan²C \n+ tanB - tanC + tanAtanBtanC - tanAtan²C + tanAtan²B - tanAtanBtanC + tan²Atan²BtanC - tan²AtanBtan²C\n+ tanC - tanA + tanAtanBtanC - tan²AtanB + tanBtan²C - tanAtanBtanC  + tanAtan²Btan²C - tan²Atan²BtanC\n= tan²AtanC - tan²BtanC - tanAtan²C + tanAtan²B - tan²AtanB + tanBtan²C\n=0 \nDivide all by tanAtanBtanC\nSo tanA/tanB - tanB/tanA - tanC/tanB + tanB/tanC - tanA/tanC - tanC/tanA = 0\n\nie tanA/tanB + tanB/tanC + tanC/tanA = tanB/tanA + tanC/tanB = tanA/tanC\n\nWhence tanA/tanB + tanB/tanC + tanC/tanA = cotA/cotB + cotB/cotC + cotC/cotA\n\nSince a triangle then tan(A + B) = tan(π - C) = - tanC\nSo (tanA + tanB)/(1 - tanAtanB) = -tanC\nie tan A + tan B = tanAtanBtanC - tanC\nso tanA + tanB + tanC = tanAtanBtanC
927315
(tanx- cotx) / (tanx+cotx) , simplify by using cotx = 1/tanx\ntanx -1/tanx)/ (tanx +1/tanx)\n=(tan^2x -1)/ (tan^2x +1)\n=(sec^2x -2)/ sec^2x ........\n[ since sec^2x = 1+ tan^2x,    so sec^2x -2 = tan^2x - 1]  :)\n=1- ( 2/sec^2x)\n= 1- 2 cos^2x\nso, (tanx- cotx) / (tanx+cotx) + 2 cos^2x =\n                     1- 2 cos^2x +2 cos^2x = 1......  :D
231477
1/3 x 1/4 = (1x1)/(3x4) = 1/12
58026
a + 1/a = 5\na = 5 - 1/a\na = 5a - 1\n4a = 1\na = 1/4
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
1150095
its sooo easy  \nsinA-sinB = 2 sin[(A-B)/2] * cos[(A-B)/2]\ncosA - cosB = 2 cos[(A-B)/2] * cos[(A-B)/2]\nthus LHS = \n(2 sin[(A-B)/2] * cos[(A-B)/2]) / (2 cos[(A-B)/2] * cos[(A-B)/2])\n= sin[(A-B)/2] / cos[(A-B)/2]\n= tan[(A-B)/2 \n= tan (A/2-B/2)\n=RHS \n\nhence proved
983394
Depends\n\nlim θ→π/2- (ie from below) tanθ = +∞\n\nlim θ→π/2+ (ie from above) tanθ = -∞ \n\nThis is because in the 1st quadrant tanθ >0 and in the second quadrant tanθ <0
850884
using chain rule and product rule\n\ny'= 3x^(2)*tan(4x)\n         +4x^(3)*sec^(2)(4x)\n  = x^2(3tan(4x)+4x*sec^(2)(4x) \n\ni hope that this helps
733318
I think your identity should be :\ncsc^2(x) / [1 + tan^2(x)] = cot^2(x)\nas you forgot to square the tan(x).\n\nWorking on the 1 + tan^2(x) gives:\n1 + sin^2(x) / cos^2(x)\n= [sin^2(x) + cos^2(x)] / cos^2(x)\n= 1 / cos^2(x)\n\nNow we have : csc^2(x) / [1 / cos^2(x)]\n= [1 / sin^2(x)] / [1 / cos^2(x)]\n= cos^2(x) / sin^2(x)\n= cot^2(x), which was to be shown.