How do I prove this? Not solving for 0, but making a proof? (1+tanx)/(1+cotx) = (tanx)?
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
784816
sin x - cos (2x) + 1 = 0\nsin x - (cos² x - sin² x) + 1 = 0\nsin x - (1 - 2 sin² x) + 1 = 0\n2 sin² x + sin x = 0\n(2 sin x + 1) sin x = 0\n\nSo either sin x = 0 or 2 sin x + 1 = 0. In the latter case:\n\n2 sin x + 1 = 0\n2 sin x = -1\nsin x = -1/2\n\nSo either sin x equals 0 (in which case x is coterminal to either 0 or π) or sin x = -1/2 (in which case x is coterminal to either 7π/6 or 11π/6). As before, we must test these solutions:\n\nsin 0 - cos 0 + 1 = 0 → 0 - 1 + 1 = 0, so x=0 is valid\nsin π - cos (2π) + 1 = 0 → 0 - 1 + 1 = 0, so x=π is valid\nsin (7π/6) - cos (14π/6) + 1 = 0 → -1/2 - 1/2 + 1 = 0, so x=7π/6 is valid\nsin (11π/6) - cos (22π/6) + 1 = 0 → -1/2 -1/2 + 1 = 0, so x=11π/6 is valid\n\nErgo, all of the solutions we just found are valid, and x is coterminal to any one of 0, π, 7π/6, of 11π/6\n\nEdit: You're welcome, Jessica.
300657
let a = b  \nMultiply both sides by a \na² = ab  \n Add (a² - 2ab) to both sides \na² + a² - 2ab = ab + a² - 2ab  \n Factor the left, and collect like terms on the right \n2(a² - ab) = a² - ab  \n Divide both sides by (a² - ab)\n2 = 1\n(The place where the proof goes wrong is the very last step.  If a² - ab = 0, then you would be dividing by zero, which is undefined.)
901670
(sin x +cos x)^2 =1    [0=<x=<360]\nsin^2x + 2sinx * cosx +cos^2x =1 or 2sinxcosx=0      Because sin^2x+cos2^x =1 \n2sinxcosx is zero when x =0, 90, 180, or 270 degrees\n\n1-sinx = cosx     [0=<x=<360]\nsinx + cosx = 1\nThis occurs when x= 0, 360
319082
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.
696270
For the first one, let's first rewrite "cscx secx" as "1 / (sinx cosx)"\n\nSo we have:\n\ntanx - (1-2cos²x)/(sinx cosx) =\n\ntanx + (2cos²x - 1)/(sinx cosx) =\n\n(sinx/cosx) + (2cos²x - 1)/(sinx cosx) =\n\n(sin²x/sinx cosx) + (2cos²x - 1)/(sinx cosx) =\n\n(sin²x + cos²x + cos²x - 1)/(sinx cosx) =\n\n(1 + cos²x - 1)/(sinx cosx) =\n\ncos²x / (sinx cosx) =\n\ncosx/sinx = cotx\n\n~ ~ ~ ~ \n\nFor the second one, you started off in the right direction!  Let's do it here:\n\n( sinx + tanx ) / (cotx + cscx) = \n\n( sinx + sinx/cosx) / (cosx/sinx + 1/sinx) = \n\n( (sinxcosx + sinx) / cosx ) * ( sinx / (1 + cosx)) =\n\n-- now I'm just going to rearrange the terms a bit --\n\n( (sinxcosx + sinx) / (1 + cosx ) * ( sinx / cosx ) =\n\n( sinx(1+cosx) / (1+cosx) ) * sinx/cosx = \n\nsinx * tanx\n\n~ ~ ~ ~\n\nTips for solving -- experience, trial and error (sorry that's not too helpful!).   What I did for the second one was I started out like you did, then I rewrote where I wanted to end up (sinx tanx) as sinx * sinx/cosx to see if I could manipulate where I had gotten so far with my problem to where I wanted to go, and I was able to see how to connect the two.  Hope that helps somewhat.  I had a hard time with trig identities when I first learned them too.  They really just take lots and lots of practice.\n\n~ ♥ ~
785851
2cos^2x+cosx-1=0\n2cos^2x+2cosx-cosx-1=0\n(2cosx-1)(cosx+1)=0\n2cosx-1=0\ncosx=1/2=cospi/3\nx=2npi+/-pi/3\ncosx+1=0\ncosx=-1=cospi\nx=2npi+-pi
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)
748572
sin(x)tan(x)=(cos(x)csc^2(x))/(cot^2(x)+cot^4(x))\nStep 1. Starting with the right side (the obviously more complicated one) factor out a cotangent squared out of the denominator.\n\n(cos x csc^2 x)/ cot^2 x (1+cot^2 x)\n\nStep 2. Use the identity csc^2 x= cot^2 x+1 in the numerator  and separate the cosine x in the top and the cotangent squared x in the bottom\n\n(cosx/cot^2 x) * (cot^2 x+1/ cot^2 x +1)\n\nStep 3. This leaves you with a cosine x divided by cotangent squared x times 1.  Use the identity that says cot x = cos x/ sin x\n\n(cos x) / (cos^2 x/ sin^2 x)\n\nStep 4. Divide the double fraction and get\n\n(sin^2 x cosx)/ (cos^2 x)\n\nStep 5. Separate your multiple factors from the squares and separate the fraction into three parts, one will be the sine x, the other will be the tangent x (because tan x = sin x/ cos x), and the cosine x divided by cosine x will go to 1.\n\nsin x * (sin x/ cos x) *(cos x/ cos x)\n\nStep 6.  Write final answer\n\nsin x tan x = sin x tan x
859187
I'll do one for u.\nIn the first,\nMultiplying both sides by x*,we get\nx+3x+2x*=7x*\n4x+2x*=7x*\nMoving 4x and 2x* to the R.H.S.\n5x*-4x+0=0\nThus 5x*-4x=0\nAttempt the others in the same manner.
496092
let x equal (tant)^2=>dx=(sect)^2dt. then simply the integrand, u will have something like integral of (cost)(sint)^2 dt. let u be sint. then u know it :D
592160
i think the answer wil be (tanX)/(tanY)\n\nSIN(X+Y)=sin x cos y+cos x sin y\n\nsin(x-y)=sin x cos y-cos x sin y\n\n use cos x cos y to divide the numerator and denominator... there is the answer...
500944
(cot(20))^2 + (cot(40))^2 + (cot(80))^2 = 9\n\nor you can think of it like this\n\n(1/tan(20))^2 + (1/tan(40))^2 + (1/tan(80))^2\n\nMultiply everything by (tan(20))^2 * (tan(40))^2 + (tan(80))^2\n\n(((tan(40)^2)(tan(80)^2)) + ((tan(20)^2)(tan(80)^2)) + ((tan(20)^2)(tan(40)^2))/((tan(20)^2)(tan(40)^2)(tan(80)^2)\n\nJust in case you don't know how to do cot\n\nor replace cot(x)^2 with csc(x)^2 - 1\n\ncsc(x)^2 = (1/sinx)^2\n\n(((1/sin(20))^2) - 1) + (((1/sin(40))^2) - 1) + (((1/sin(80))^2) - 1)\n\n((1/sin(20))^2) - 1 + ((1/sin(40))^2) - 1 + ((1/sin(80))^2) - 1\n\n((1/sin(20))^2) + ((1/sin(40))^2) + ((1/sin(80))^2) - 3\n\n((1/sin(20))^2) + ((1/sin(40))^2) + ((1/sin(80))^2) = 12\n\n12 - 3 = 9
301417
I'll try. I gave this proof to THREE people already:\n\nTake any two ordinary constants. (e.g. x and y)\nMake them alike, as in x = y = any non-zero number (in this case, 2)\nMultiply each by y, but x works equally well (y²=xy).\nSubtract x² from each equation (y²-x² = xy-x²).\nFactor y²-x² into (y+x)(y-x) to make ((y-x)(y+x) = xy-x²).\nFactor xy-x² into x(y-x) to make ((y-x)(y+x) = x(y-x)).\nDivide by (y-x) to get (y+x=x).\nSubtract x from both sides of this equation to get y=0.\nThis means that y=0.\n\nbut since, y equaled 2 in the first case, that means 2 = 0.\nIf you multiply this equation by 2.5, you get 5 = 0.\n\nSince 0 + 0 = 0, you can plug in 2 for the first two zeroes, and five for the last one, and then, TA DA! 2 + 2 = 5.\n\nP.S., this proof has a little falsity in it. you can't divide by (y-x) because that is zero!
932480
sin(A+B)sin(A-B)/(cos²Acos²B)\n= (sinAcosB + cosAsinB)(sinAcosB - cosAsinB)/(cos²Acos²B)\n= (sin²Acos²B - cos²Asin²B)/(cos²Acos²B) (since (x + y)(x - y) = x² - y²)\n= sin²A/cos²A - sin²B/cos²B\n= tan²A - tan²B ← Option c)\n\nPlease note: sin(A + B) = sinAcosB + cosAsinB\nand sin(A - B) = sinAcosB - cosAsinB
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.
932836
Are you sure this is the correct identity?\n\n 1+ tanx.tan2x = 1 + (sinx/cosx)(sin2x/cos2x)\n                        = 1 + (sinx/cosx)(2sinxcosx/cos2x)\n                        = 1 + 2sinx.sinx/cos2x\n\n cos2x = 2cosx.cosx - 1  = 2cos^2x - 1        (Note:  ^2 = square)\n\n then..\n      1+tanx.tan2x = 1 + [2sin^2x/(2cos^2x - 1)]\n                            = (2cos^2x + 2sin^x -1) / (2cos^2x - 1)\n                            = [2(cos^2x + sin^2x) -1] / (2cos^2x - 1)\n                            = 1/cos2x    (because 2cos^2x - 1 = cos2x)\n\n     Then 1/1+tanxtan2x = cos2x