In a triangle ABC tanA+tanB+tanC=?
447477
In a triangle the sum of the angles is 180°, but the sum of the tangents is not anything in particular.  The tangent function is 0 for 0°, but it increases without bound as the angle approaches 90°.  So if the triangle contains one angle very very close to 90° and the other 2 are the rest of the 180°, the sum of tangents can be as large as you like.  On the other hand, if two of the angles are very very close to 0 and the third very close to 180, the sum of the tangents can be as close to 0 as you like.
72362
I forgot how to do those, but it's some sort of trig identity isn't it?  Like\n\nsqrt (1-tan [x])\n\nProbably not though.
291250
Factor the left side: 2cosx(tanx+1)=(tanx + 1)\n\nThis is true whenever tanx+1 = 0, or 2cosx = 1\n\ntanx = -1 in degrees, -135 or 315\ncosx=1/2 in degrees, 60 or 300\n\n(or multiples of 360 plus the above)
815598
a+b+ c+ab+cb+ac = abc + 1\nor a+b+c - abc = 1- ab-cb-ac\nor (a+b+c - abc)/(1-ab-bc-ac) = 1\n\nputting a = TAN A , b= tan B, c = tan C\n\nwe get LHS = tan(A+B+C) = 1\nor A+B+C = tan ^-1 = pi/4+npi\n\n chose any a and b then c can be chosen as\n\ntan^1(pi/4 - tan A - tan B)
155419
the base of a triangle times the height times 2\n\nb*h*2
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
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)
528735
tan by definition is equal to sine/cosine.\n\nProof:\n\ntan = opp/adj\nsin = opp/hyp\ncos = adj/hyp\n\nsin/cos = (opp/hyp)/(adj/hyp) \n= opp/hyp*hyp/adj \n= opp/adj\n= tan\n\nSorry about that, knew that was too simple :P!\n\nNEW ANSWER:   You nedd to draw out the triangle and solve from there.  Let there be a triangle with adjacent side a,  and opp side b.  \n\nThen if tanbeta = 2, opp= 2, adj = 1 and hyp = sqrt(5)   from the pythagorean theorem.\n\nThus cos = 1/sqrt(5) and sin = 2/sqrt(5)\n\nThe quadrant only tells you the sign of cos and sin or how the angle is measured.  Since you're in QI, both sin and cos are positive so no modification to the above answers are needed.\n\nNow you might wonder why I said opp was 2 and adj was 1.  It doesn't matter (can't remember the proof for that right now).  As long as you pick a solution that works, say 4 and 1 or 3 and 1.5, it will ALWAYS simplify to the answer I just gave you.
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
857149
we know\n\ntan A/2 = sin A/(1+cos A) = sinA (1- cos A)/(1- cos^2 A)\n= (1- cos A)/ sin A = cosec A - cot A\n\nusing this we get cot A = cosec A - tan A/2\n\n2) let( A+B)/ 2 = x\n(A-B)/2 = y\n\nA = (x+Y)\nB= x-y\nsin A + sin B = sin (x+y) + sin (x-y) = 2 sin x cos y\nsin A - sin B = sin (x+y) - sin (x-y) = 2 sin y cos x\n(sin A+ sin B)/(sin A - SIn B) = tan x/tan y = tan (A+B)/2/tan(A-B)/2\nQED
1023518
Use the Law of Sines.\n\nsin A/a = sin B/b\n(sin 36)/24 = sin B/47\nsin B = (47 sin 36)/24\nsin B = 1.151079452\nAt this point you have a problem--it is impossible for the sine of an angle to have a value greater than 1.\n\nSo either you copied the problem wrong, or you are being asked to recognize that a triangle cannot be formed with these dimensions.
905611
If u r saying that u have three sides of a triangle as a,b,c\nAnd u want to determine angle b/w them, u can use the formula\ncosA=(b^2+c^2-a^2)/2bc where A is the angle formed by the sides b and c.similarly if B is the angle formed by a and c, it can be calculated as\ncosB=(a^2+c^2-b^2)/2ac\nonce u have calculated A and B, subtract their sum from 180, u will get third angle.\nRemember u will have to use inverse table for cos to find A and B
139883
(a+b)(a-b) \n= a^2 - ab + ba - b^2\n= a^2 - b^2 -ab +ab\n= a^2 - b^2
104855
There is many answers\na=1 b=2 c=3 is true\nalso a=b=c=sqrt(3) is true but not integers\n\nHow we can get them? \na,b,c must satisfy abc=a+b+c\n\nWhy?\natan(a) + atan(b) + atan (c) =pi\nlet atan(b)+atan(c) = y     (eq1)\n\natan(a) + y =pi\ntaking tan to the 2 sides\n\ntan [atan(a) + y] =tan(pi) = 0\n[tan(atan(a)) +tan (y) ]/[1- tan(atan(a))*tan (y) ]=0\n\nso a + tan(y)=0                       (eqn2)\nbut tan(y) = tan (atan (b) +atan(c))    from eqn1\n=(b+c)/(1-bc)\n\nsubstitue in eqn2\na+(b+c)/(1-bc)=0\nmultiply by (1-bc)\na(1-bc)+b+c=0\na+b+c-abc=0\n***********\n*a+b+c=abc*\n***********\n\na=1 b=2 c=3 is true since 1+2+3 =1*2*3 =6\na=b=c=sqrt(3) is true since 3sqrt(3)=srqt(3)^3
497748
In rt triangle ABC with B= 90degrees (draw figure)      \nIf C is theta, cosC=7/8(ignoring sign)\ni.e., BC = 7, CA = 8 (in simplest terms)\nusing pythagoras theorem, AB= sqrt(15)\n \nsince C=theta is in the second quadrant tan C will be negative\n\nso tan C = AB / BC \n\n  tan C= - { sqrt(15) } / 7\n\ntan(theta) = -0.5533
716883
Trying for wow #2---\n\nR = .15 m\nω = 2π/T = 13.963 rad/sec\nA little vector diagram shows that tanθ = rω²m/mg = rω²/g\nbut r = Rsinθ, so tanθ = sinθ/cosθ = Rsinθω²/g  from which\n\n1/cosθ = secθ = Rω²/g = .15*13.963²/9.8 = 2.984 → θ = 70.42°\n\nYou get to do b)\nGood luck..
247800
it is basically:\na2 + b2 = c2   -->  (A squared)+(B squared)=(C squared)\n  \n\nif you have a right triangle.you have three sides.the two legs.and the hypotenus(sp?).a and b are the two legs and c is the hypotenus.\n\nin an equation.if you are given a and b.you can square each of them and then add the two values together.the will give you c squared.to find c you then need to take the square root of the value(a squared + b squared).\n\nif you have c and a leg(example:a).you square c.and then subtract a squared from that value(c squared).that will give you b squared.to get b you just get the square root of c squared - a squared.\n\n\nhope that helps.