tanA+tanB+tanC-tanAtanBtanC over 1-tanAtanB-tanBtanC-tanCtanA= tanA + B + C?
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)
672272
I have taken the liberty of calling it the (a+b)^2=a^2+b^2+2ab ; (a+b+c)^2=a^2+b^2+c^2+2ab+2ac+... ; (a+b+c+d)^2=a^2+b^2+c^2+d^2+2a...\n+2cd etc... theorem. A bit long winded but exact.
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
880634
tan²x =\n\nsin²x\n------- =\ncos²x\n\n½(1 - cos2x)\n------------------ =\n½(1 + cos2x)\n\n1 - cos2x\n------------- =\n1 + cos2x\n\n2 - 1 - cos2x\n----------------- =\n1 + cos2x\n\n2 - (1 + cos2x)\n-------------------- =\n1 + cos2x\n\n2\n-------------  - 1\n1 + cos2x\n\nOR\n\ntan²x =\n\nsec²x - 1 =\n\n1\n--------  - 1 = \n cos²x\n\n1\n------------------  - 1 =\n½(1 + cos2x)\n\n2\n-------------  - 1\n1 + cos2x
756689
simple\n\nusing the fomula tan(A+B) = (tan A + tan B)/(1-Tan A tan B)\nputting A = pi/4\n\ntan (pi/4 + B) = (1+ tan B)/(1- tan B) \n= (1+ sin B/cos B)/(1- sin B/cos B) \n= (cos B-sin B)/(cos B + sin B)\nput b = x/2 to get the result
139883
(a+b)(a-b) \n= a^2 - ab + ba - b^2\n= a^2 - b^2 -ab +ab\n= a^2 - b^2
1144353
4a+4b+4c=2ab+2bc+2ac \n4(a+b+c)=2(ab+bc+ac)\na=2\nb=2\nc=2\n\nor \na=1\nb=2\nc=4
1162151
Use the fact that tan^2(x) = sec^2(x ) - 1 and integrate that.  You will get tanx -x +c.
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.
841196
This is interesting....\n\nTry this.  First, change all tan(x) and cot(x) into sin(x)/cos(x) and cos(x)/sin(x) respectively.  Then combine the fractions using common denominator which is cos(x)sin(x).\n\nEventually, the fraction will reduce to sin^2(x) -cos^2(x)\n\nPlug this back into the original equation and you get:\nsin^2(x) - cos^2(x) + 2cos^2(x) = 1\nsimply and you get:\nsin^2(x) + cos^2(x) = 1\n1 = 1\n\nX can be any real number.....
564798
Ba(NO3)2 + Na2CO3-->BaCO3 +2NaNO3\n\nto do a double displacement follow this formula\n\nAB + CD --> AD + CB
119175
You will need to use the formula for tan(a+b) and the formula for tan(2a).\n\nrewrite tan (3a) as (tan(2a +a))\n\nthis becomes (and I hope I can type it)\n\n[tan (2a) + tan(a)]  all over [1 - tan(2a)tan(a)]\n\nnow replace all the tan (2a)'s with 2tan(a)/(1-tan^2 a)\nand you get a huge complex fraction\n\nthe common denominator for the fractions in the complex fraction is 1 - tan^2 a.  Get this in all 4 denominators & cancel it out, leaving\n\n[2 tan(a) + tan(a) - tan^3(a)] over [1 - tan^2(a) - 2tan^2(a)]\n\nthis simplifies to [3 tan(a) - tan^3 (a)] over [1 - 3 tan^2 (a)]\n\nThe last part of your problem got chopped off; hope this is right.
743138
join AC\nnow youhavetwo triangles ABC and ADC\nin triangle ABC <CAB+<B+<ACB=180*.............(1)<s of a triangle\nin triangle ADC <CAD+<D+<ACD=180*............(2)--------do--------\nadding (1) and (2)\n<ABC+<ADC+<B+<D+ACB+<ACD=360*\n=><DAB+<B+<BCD+<D=360*\nhence proved
900488
cos(x) +sin(x)tan(x)=sec(x)\nmultiply by cos(x) on both sides\ncos^2(x)+sin^2(x)=1\n1=1\nI believe they are probably asking you to prove the identity
8794
5t+1>3t-13\n-> 5t-3t>-13-1\n-> 2t>-14\n-> t>-14/2\n-> t>-7\nKeep Asking
970698
(a + b)(c + d + e)\n= a(c + d + e) + b(c + d + e)\n= ac + ad + ae + bc + bd + be\n\nOR\n\n(c + d + e)(a + b)\n= c(a + b) + d(a + b) + e(a + b)\n= etc.
1093073
A = 45 degrees or (pi / 4) radians\n\n1/sin^2A-1/cos^2A-1/tan^2A-1/cot^2A-1/sec^2A-1/csc^2A = -3\nIf i understand the way you wrote the problem above, it's 1/ each term, subtract each other right?\n1/ ((sinA)^2) - 1/ ((cosA)^2) - 1/ ((tanA)^2) - 1/ ((cotA)^2) - 1/ ((secA)^2) - 1/ ((cscA)^2) = -3\nI'm going to drop the A part, i assume you know that it will still be officially added in the final version\n\n[1/ (sin^2)] + [(-1)/ (cos^2)] + [-(cot^2)] + [-(tan^2)] + [-(cos^2)] +  [-(sin^2)] = -3\n[1/ (sin^2)] + [(-1)/ (cos^2)] + [-(cos^2) / (sin^2)] + [-(sin^2) / (cos^2)] + [-(cos^2) - (sin^2)] = -3\ncombine denominators\n[( 1 - (cos^2) ) / (sin^2)] + [( (-1) - (sin^2) ) / (cos^2)] + (-1) [(cos^2) + (sin^2)] = -3\n[(sin^2) / (sin^2)] + (-1)[( 1 + (sin^2) ) / (cos^2)] + (-1) [ 1 ] = -3\nTrig identity [(cos^2) + (sin^2)] = 1\n[ 1 ] + (-1)[( 1 + (sin^2) ) / (cos^2)] + [ -1 ] = -3\n(-1)[( 1 + (sin^2) ) / (cos^2)] = -3\n[( 1 + (sin^2) ) / (cos^2)] = +3\n1 + (sin^2) = 3 * (cos^2)\nTrig identity 1 = [(sin^2) + (cos^2)] =\n[(sin^2) + (cos^2)] + (sin^2) = 3 * (cos^2)\n2 * (sin^2) = 2 * (cos^2)\nsin^2 = cos^2\nsinA = cosA\nA= 45 degrees of (pi/4) radians