given the formula tan(a-b)=tamA-tanB/1+tanAtanB?
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
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)).
858246
Interpretation # 1\n\n2/3 tan x - 2 tan² x =1\n\nso 6 tan² x - 2 tanx + 3 = 0\n\nDiscriminant < 0 so no real solutions\n\nInterpretation # 2\n\n2 tan (x/3) - 2 tan² x =1\n\ntan 3t = tan (t + 2t)\n= (tan t + tan 2t)/(1 - tan t tan 2t)\n= (tan t + 2tan t/1 - tan² t)/( 1 - tan t*2tan t/1 - tan² t)\n= (tan t (1 - tan² t) + 2tan t)/ (1 - tan² t - 2tan² t)\n= (3tan t - tan³ t)/(1 - 3tan² t)\n\nLet tan x/3 = p Then tan x = p(3 - p²)/(1 - 3p²)\nSo 2 tan (x/3) - 2 tan² x\n= 2p - 2 (p(3 - p²)/(1 - 3p²))²\n\n= [2p(1 - 3p²)² - 2p²(3 - p²)²]/(1 - 3p²)² = 1\nSo 2p[(1 - 3p²)² - p(3 - p²)²] = (1 - 3p²)²\nie (2p - 1)(1 - 3p²)² - 2p²(3 - p²)² = 0\nSo (2p - 1)(1 - 6p² + 9p^4) - 2p²(9 - 6p² + p^4) = 0\n\nSo 18p^5 - 9p^4 - 12p³  + 6p² + 2p - 1 - 2p^6 + 12p^4 - 18p² = 0\nie 2p^6 - 18p^5 - 3p^4 + 12p³  + 12p² - 2p + 1 = 0\n\ntan(x/3) ≈ 1.09073, 9.08455 (by equation solver)\nSince 0 ≤ x < 2π then 0 ≤ x/3 < 2π/3\n\nSo x/3 ≈ 0.82877, 1.46116\nSo x ≈ 2.4863, 4.3835
1162151
Use the fact that tan^2(x) = sec^2(x ) - 1 and integrate that.  You will get tanx -x +c.
960686
The equation is correct but only if A is 0, 180 or integer multiples of 180 degrees.\nsecA/(1+sinA) - cosA/(1+sinA) = tanA\n(secA - cosA)/(1+sinA) = tanA\ncosA(secA - cosA)/(1+sinA) = tanA * cosA\n(1-cos^2A)/(1+sinA) = sinA\nsin^2A/(1+sinA) = sinA\nsin^2A = sinA + sin^2A\nsinA = 0  which occurs when A=0,180,360,etc.
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...
881848
First apply the identities tan x = sin x/cos x and cot x = cos x/sin x:\n\n(tan x - cot x)/(tan x + cot x) + 2 cos^2 x = (sin x/cos x - cos x/sinx)/(sin x/cos x + cos x/sin x) + 2 cos^2 x.\n\nNow get a common denominator and combine some fractions:\n\n(sin x/cos x - cos x/sinx)/(sin x/cos x + cos x/sin x) + 2 cos^2 x =\n((sin^2 x - cos^2 x)/ sin x cos x)/((sin^2 x + cos^2 x)/ sin x cos x) + 2 cos^2 x\n\nWe can cancel the sin x cos x's on bottom, and get\n\n((sin^2 x - cos^2 x)/ sin x cos x)/((sin^2 x + cos^2 x)/ sin x cos x) + 2 cos^2 x = (sin^2 x - cos^2 x)/(sin^2 x + cos^2 x) - 2 cos^2 x.\n\nNow, applying the identity sin^2 x + cos^2 x = 1 gives\n\n(sin^2 x - cos^2 x)/(sin^2 x + cos^2 x) - 2 cos^2 x = (sin^2 x - cos^2 x) + 2 cos^2 x.\n\nSimplifying gives\n\n(sin^2 x - cos^2 x) + 2 cos^2 x = sin^2 x + cos^2 x.\n\nAnd finally\n\nsin^2 x + cos^2 x = 1.
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
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
413333
let suppose that \nr(@) = 4/ sqrt( 4cos^2(@) +sin^2(@) )\nthen x(@(t) ) = r(@) . cos (@) emplies\n\nx^2 (t) = r^2 (@) . cos^2(@)\n\nx^2 = 16 / (4 cos^2 + sin^2)  . cos^2\n\nso x^2(t) = 16/ (4+ tan^2(t) )\n\nBut tan( @(t) ) = y(t) / x(t)\n\nThus: x^2 = 16 / (4+ y^2 / x^2 )\nThus: \nEQ : x^2/4 + y^2/16 =  1\n\nCheck @=0  --> r(@=0) = 2\nAnd (x=2 , y=0 ) is solution of EQ
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.
756636
tan² x - 2 tan x = 0\n\nFactor out tan x\ntan x(tan x - 2) = 0\n\nTherefore,\ntan x = 0, or tan x - 2 = 0\n\nThus,\nx = arctan 0 or arctan 2\n\n^_^
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
938747
You have sin (arctan( -5 / 7 ))\n\n(tan-1 is the same as arctan, the angle with tangent -5/7).\n\nSOH-CAH-TOA\n\ntan(angle) = -5/7\ntan(angle) = opposite/adjacent\nopposite = -5\nadjacent = 7\nhypotenuse = sqrt(25 + 49)\nhypotenuse = sqrt(74)\n\nsin(angle) = opposite/hypotenuse\nsin(angle) = -5/sqrt(74)
676590
Sorry, I can't see the photo...but:\n\nSOH - sin = opp/hyp\nCAH - cos = adj/hyp\nTOA - tan = opp/adj
1026215
ok i got the answer to this\ni just solved it now and here's the solution.if u have time and patience then u can go through it ........\n\ntan a{tan (a+60)} + tan a {tan (a -60)} + tan (a+60)tan(a-60)\n\n= tan a(tan a + tan60) / (1-tan a tan60) + tan a (tan a - tan 60) / (1+tan a tan60) + {(tan a + tan 60) / (1- tan a tan 60) } {(tan a -tan 60)/1 + tan a tan 60)}\n\nnote ^ is to the power of,in case u didn't know.and one more thing,make sure u know all the formulae used in this problem.this will help u understand my steps better! and sqrt = sqaure root of .\n\n=( tan^2 a+ tan a sqrt3) / (1 - sqrt3tan a) + (tan ^2a - sqrt 3 tan a )/(1 + sqrt 3 tan a ) + (tan ^2a - 3)/(1 - 3 tan^2 a )--- (this particular thing (tan ^2a - 3)/(1 - 3 tan^2 a ) i got it using the formula (a+b)(a-b)=a^2-b^2 after mulipying {(tan a + tan 60) / (1- tan a tan 60) } {(tan a -tan 60)/1 + tan a tan 60)}\n\nand now that becomes\n= {(1 + sqrt3 tan a )(tan ^2 a+ sqrt 3tana) + (1 - sqrt 3 tan a)(tan^2 a - sqrt3 tan a) + tan^2 a -3} / {1 - 3tan^2a}\n=( tan^2a + sqrt3tan^3a + sqrt 3 tana + 3tan^2a + tan^2a - sqrt 3tan ^3 a - sqrt3tan a + 3tan a  + tan^2 a -3) / ((1 - 3 tan^2 a )\n\nnow u can cancel out + sqrt3tan^3a  and - sqrt3tan^3a ,+sqrt 3 tan a and -sqrt 3 tana .\n\nwhich becomes\n= (9tan^2a - 3) / (1 - 3tan^2a)\n= -3(-3tan^2a + 1) / (1 - 3tan^2 a) \n= -3\n\n\nall the best for your studies dear friend.
935235
.. lim ____x____ =\nx →0 tan^-1 (4x)\n\n.. lim _____1_____ =\nx →0  4/[1 + (4x)²]\n\n.. lim _1_+_(4x)²_ =\nx →0 ...... 4\n\n= 1/4