if tan(a+b)=3 and tanb=1/7, find tana please show work?
443111
"god" is not correct...the problem is not that simple; however, it is not too hard\nHere's the formula you need:\ntan(A+B) = (tan A + tan B) / (1 − tan A tan B)\nPlug in 1/7 for tanB and 3 for tan(A+B) (and just to make the typing simpler I'm replacing tanA with x)\n3 = (x + 1/7)/(1 - x(1/7))\nMultiply both sides by 1 - x(1/7)\n3(1 - x(1/7)) = x + 1/7\nDistribute the 3\n3 - (3/7)x = x + 1/7\nAdd (3/7)x to each side \n3 = x + (3/7)x + 1/7\nCombine x + (3/7)x = (7/7)x + (3/7)x = (10/7)x\n3 = (10/7)x + 1/7\nSubtract 1/7 from each side\n3 - 1/7 = (10/7)x\n21/7 - 1/7 = (10/7)x\n20/7 = (10/7)x\nMultiply by the reciprocal of 10/7\n(20/7)(7/10) = x\nx = 2\nThus, tanA = 2
831486
Differentiate both sides, we have \n\ndy =  (1/x)dx + (1 + tan^2(y))/tany dy\n\ntherefore\n(1 - (1 + tan^2(y))/tany)dy=(1/x)dx,\n\ni.e.\n\ndy/dx=tany/(x(tany - tan^2- 1)).
885632
Yes you are correct.\n\nx=tan t = (sin t)/(cos t)\nso cos t = (sin t)/x\nSo y^2 = (sin^2 t)/x^2 = (1 - cos^2 t)/x^2 = (1 - y)/x^2\n\nRearranging this you get y = 1/(x^2 +1)
886349
SOHCAHTOA \nSOH =sine=Opposite /Hypotenuse\nCAH cos=adjacent/hypotenuse\nTOA tan=opposite/adjacent\nIn finding the adjacent,we use CAH\ncos19=adjacent/13.8\nadjacent=cos19x13.8\nadjacent=13.048
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
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
932500
I don't know any good math websites but I do know how to solve this.  b would be the side adjacent to the 40 degree angle and a is the side opposite it.\n\nTangent = opposite leg/adjacent leg so\ntan 40 = 15/b\n\nb tan 40 = 15\n\nb = 15 / tan 40
746463
I'm guessing the second one should be h/(x+300), not h/x + 300.\nLets make h the subject of each.\nh = x tan 31\nh = (x+300) tan 24.\nEquate those:\nx tan 31 = (x+300) tan 24\nx tan 31 = x tan 24 + 300 tan 24\nx(tan 31 - tan 24) = 300 tan 24\nx = 300 tan 24 / (tan 31 - tan 24).\nYou can then work out h = x tan 31.\nYou should get x = 858.2, h = 515.7, approx.
1023705
2 + 2 / 2 =\n2 + 1 = 3
921171
If tan( t ) = -2 then t = atan( -2 ) = -1.107 + multiples of π.\n\nIn the range from 0 to 2π the required angles are\n\n-1.107 + π and -1.107 + 2π, or\n\n2.034 and 5.176
1050478
tan(x) = y  implies arctan(y) = x\n\narctan ( x )  = tan-1 ( x )\n\ndoes this help u
730760
θ = α + β  where α = tan^-1 (1/3) and β = tan^-1 (1/2)\nie tanα = 1/3 and tanβ = 1/2\n\ntanθ = tan(α + β)\n= (tanα + tanβ)/(1 - tanαtanβ)\n= (1/3 + 1/2)/(1 - 1/3 * 1/2)\n= (5/6)/(5/6)\n= 1\nSo θ = ....., π/4, 5π/4, .....\nSo cosθ = ±1/√2 = ±√2/2
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
549518
How about...\n\n( 3 + (3/7) ) * 7 = 24\n\nThat would be like (21/7 + 3/7)... which is (24/7)... times 7 = 24.
868863
tan^2x+sec^2x-2tanxsecx\n=(sin^2x/cos^2x)+(1/cos^2x)-2(sinx/cosx)(1/cosx)\n=(sin^2x+1+2sinx/cos^2x\n=(1+sinx)^2/(1-sin^2x)\n=(1+sinx)/(1-sinx)
764150
I presume you want to simplify it, though you didn't say that :P\nWith any trig expression like this, you have two options:\n1) Look for some cute trick. This is usually what you are shown when you get given the answer. However, these are often tricky to find.\n2) Turn everything into sin and cos, and simplify. Then use the fact that (sin x)^2 + (cos x)^2 = 1. This *always* works!\n\nSo, turn tan into sin/cos, and sec into 1/cos:\n((sin B)^2 - (sin B)^2/(cos B)^2) / (1 - 1/(cos B)^2)\nNow, multiply top and bottom by (cos B)^2:\n((sin B)^2 * (cos B)^2 - (sin B)^2) / ((cos B)^2 - 1).\nNow, factorise the top:\n(sin B)^2 ((cos B)^2 - 1) / ((cos B)^2 - 1)\nAnd cancel:\n(sin B)^2.
310146
i think u can do this:\n               \n            (secA+tanA)2\n   =sec2A-2tanA.secA+tan2A\n  now,   put the value of secA that can b derived from the question\n             secA+tanA=2\n       or,secA=2-tanA\n                =sec2A-2tanA.2-tanA+tan2A\n           =sec2A-4tanA-2tan2A+tan2A\n             =sec2A-4tanA-tan2A\n              =tan2A-1-4tanA-tan2A\n             =-(1+4tanA)\n\nthis may b the ans(in the above solution in some places 2 means square,i hope u'll understand it)