How would I use trig identities to verify the following:\n\n(cot x)/(1-tan x) + (tan x)/(1-cot x) = 1 + tan x + cot x\n\nShowing the steps involved would be greatly appreciated.\nThank you very much for your help!
121130
Oh yes i have cracked it at last....\n\nsolution...\n\nLet x=tan-1 (y) (i mean tan inverse)\nthen tan x = y,and cot x = 1/y (coz tan-1 (x) = cot-1 (1/x))\n\nthen ur given equation's Left hand side part (LHS) will be\n\n(1/y)/(1-y) + (y)/(1-(1/y))\n\n= 1/(y-y2)+ y/(1-(1/y))\n\n= 1/(y-y2) + y2/(y-1)\n\n= y2(y-1) - 1/y(y-1)\n\n= y3/y(y-1)\n\nI hope u understood till this step.\n\nNow we have (a3-b3) = (a-b) (a2+ab+b2)\n\nthen the LHS becomes...\n\n(y-1)(y2+y+1)/y(y-1)\n\n= (y2+y+1)/y\n\n= y+1+(1/y)\n\nsubstitute back y=tanx, 1/y = cot x and leave 1 as it is...\n\nU have LHS = 1 + tan x + cot x...Thats ur RHS !!!
500709
it's "pi" not "pie" dear ... :-)\n\nAnd, yes I do know how to prove those identities.\n\nAnd, no I won't do them all, as it will deprive you of the wonderful opportunity to learn and practice for your test. \n\nBut, I will do one for you!  I'll do #8:\n\n(sinx + cosx) ² =\nsin²x + 2sinxcosx + cos²x = \nsin²x + cos²x + 2sinxcosx = \n1 + 2sinxcosx  {{ using trig ID #1 }} = \n1 + sin2x {{ using trig ID #2 }}\n. . . . .\nTrig ID #1 :  sin²x + cos²x = 1\nTrig ID #2 :  sin2x = 2sinxcosx\n\nNRT
366796
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.
108902
cot = o/a\ncsc = r/o\n\no = opposite side\na = adjacent side\nr = radius\n\ncalculate csc(2x) in terms of o and r.\nSubstitute o/a for cot(x) and substitue the value you determine for csc(2x) \nmultiply these, then convert to a trig function in terms of x.\n\nIf you know trig identities better than I do, you can reduce this expression using identities, without resorting to o, a, and r.  However, either route will work.
459448
Hi, on your first one,\nsin^3 θ cos θ - sin θ cos^3 θ = -1/4 \n\nFactor out sin θ cos θ (sin^2 θ - cos^2 θ) = -1/4\n You will notice that if the part in parentheses were reversed it would be an identity for cos 2θ so multiply both sides by -1 to have this happen:\n\nsinθ cosθ (cos^2θ - sin^2θ) = 1/4\n\nYou will also notice if there were a 2 in front you'd have an identity for sin 2θ so multiply both sides by 2 to get\n\n2 sinθcosθ (cos^2θ - sin^2θ) = 1/2\n\nreplace with the identities: sin 2θ cos2θ = 1/2\n\nNow, if this had a 2 in front it would be an identiti for sin 2(2θ) so mult. both sides by 2 to get\n2 sin 2θ cos 2θ = 1\n\nsin 2(2θ) = 1\n\nsin 4θ = 1\n\nYou know that sin pi/2 = 1 so 4θ = pi/2 and θ = pi/8.  You can figure the additional answers as θ increases.\n \n\n If I get the second I'll edit this unless someone else already got it\n\nHere's what I think:\nsec 4θ + 2 sin 4θ = 0\n\n1/cos4θ + 2 sin 4θ = 0\n\nget common denominators\n\n1/cos 4θ + 2 sin4θ cos4θ/cos 4θ\n\nyou will see that the top of the second fraction is an identity for sin 2(4θ) so replace it\n\n1/cos4θ + sin 2(4θ)/cos4θ = 0\n\nThis is 1 + sin 8θ all over cos 4θ = 0\n\nIf a fraction = 0, its numerator = 0 so 1 + sin 8θ = 0\n\nsin 8θ = -1\n\nsin of 3pi/2 = -1 so 8θ = 3pi/2 and θ = 3pi/16\n\nagain there are infinitely many more answers unless you were given a specific domain for θ\n\nHope that helps
362293
tan x + tan 2x + tan 3x = 0\nor, tan x + tan 2x = -tan 3x\nor, sin x/cos x + sin 2x/cos 2x = -sin 3x/cos 3x\nor, (sin x *cos 2x + cos x * sin 2x )/cos x * cos 2x =  -sin 3x/cos 3x\nor, sin (2x + x )*cos 3x = - cos x * cos 2x * sin 3x\nor, sin 3x * cos 3x +  cos x * cos 2x * sin 3x = 0\nor, sin 3x (cos 3x + cos x * cos 2x)= 0\nor, sin 3x [cos ( 2x + x ) + cos x * cos 2x)= 0\nor, sin 3x [ cos x * cos 2x - sin x * sin 2x+ cos x * cos 2x)= 0\nor,  - sin 3x * sin x * sin 2x = 0\nor, sin x * sin 2x * sin 3x = 0\n\neither, sin 3x = 0\nie, 3x = nπ , n ε I\nie, x =  (nπ/3) , n ε I\n\nor, sin 2x = 0\nie, 2x =  nπ , n ε I\nie, x =  (nπ/2) , n ε I\n\nor, sin x = 0\nie,  nπ , n ε I\n\nBut putting x = nπ/2 , does not satisfy the equation\n(eg. Putting x = π/2, tan π/2 = undefined)\n& the solution set x = nπ is already included in the set x = nπ/3\nso the required solution is , x = nπ/3\n\nhence the general solutions of x are, x =  (nπ/3) n ε I ie. n = ....-3,-2,-1,0,1,2,3.......\n\n\nP.s. don't those dummies know about the significance of a G.S.\nI,m not telling it to you, you may make out the cause of such comment if you schroll down.
355582
(sec(x)^4)+(tan(x)^4)\n=((sec(x)^2)+(tan(x)^2))*((sec(x)^2)-(tan(x)^2))\n=(sec(x)^2)+(tan(x)^2)
198876
(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)
487465
This may be unexpected, but a good site with the formulas you need.\n\nhttp://en.wikipedia.org/wiki/Trig
173778
sin(x) - cos(x) - tan(x) = -1\nsin(x) - cos(x) - ((sin(x))/(cos(x))) = -1\n(sin(x)cos(x)) - cos(x)^2 - sin(x) = -cos(x)\n(sin(x)cos(x)) - sin(x) = cos(x)^2 - cos(x)\n((sin(x))(cos(x) - 1) = (cos(x))(cox(x) - 1)\n\ndivide both sides by (cos(x) - 1)\n\nsin(x) = cos(x)\nor\nsin(x) = 1cos(x)\n\ndivide both sides by cos(x)\n\n(sin(x))/(cos(x)) = 1\n\ntan(x) = 1\n\ninverse tan both sides\n\nx = 45° or 225°\n\nAlso i found out that "x" also equals 0\n\nIf you go to www.quickmath.com, it will give you answers of (pi/4) which equals 45°, (-3pi/4) which equals 225°, and 0.
517195
= cosx[cosxcosy+sinxsiny]+sinx[sinxcosy-cosxsiny]\n\n= (Cosx)^2cosy+cosxsinxsiny+(sinx)^2cosy-cosxsinxsiny\n\n= (Cosx)^2cosy + (Sinx)^2cosy\n\n= Cosy [ (cos)x^2 + (sinx)^2]\n\n= Cos(y)
395836
cot²(x)\n------------ =  ............ (NOTE cot²x + 1 = csc²x)\ncsc(x) - 1\n\ncsc²(x) -1\n------------ =\ncsc(x) - 1\n\n(csc(x) - 1)(csc(x) + 1)\n----------------------------- =\ncsc(x) - 1\n\ncsc(x) + 1
554584
The two identities that are important for this equation are:\n\n(sin x)/(cos x) = tan x\n\n(sin x)^2 + (cos x)^2 = 1\n\nI found it easiest to work on the right side of the equation, and figure it in terms of sin x.\n\ncos x / tan x\n= cos x / (sin x / cos x)\n= cos x * (cos x / sin x)\n= (cos x)^2 / sin x\n\nNext, the (cos x)^2 is equal to 1 - (sin x)^2, so substituting, we get:\n\n= (1 - (sin x)^2) / sin x\n\nAdd finally, splitting the numerator,\n\n= 1/sin x - (sin x)^2 / sin x\n= 1/sin x - sin x\n\nWhich is the left side of the equation.  This equation is therefore valid for all non-zero values of sin x.
444144
(1-sin(t))/cos(t)\n=[(1-sin(t))(1+sin(t))]/[(1+sin(t))cos(t)]\n=(1-(sin(t)^2))/[(1+sin(t))cos(t)]\n=(cos(t)^2)/[(1+sin(t))cos(t)]\n=cos(t)/(1+sin(t))
346551
tan(a+b)=sin(a+b)/cos(a+b)=\n(sinacosb+sinbcosa)/\n(cosacosb-sinasinb)\ndivide top and bottom by cosacosb\n(tana+tanb)/(1-tanatanb)
458963
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
134711
RHS cot a - tan b \n= cos a/sina - sinb/cosb \n= cosacosb - sinasinb/sinacosb\n= cos(a+b)/sinacosb\n= LHS (proven)\n\nThe idea is to prove that both sides are the same.