(tan x- cot x)/( tan x + cot x)+ 2cos^2 x=1?
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
310146
i think u can do this:\n               \n            (secA+tanA)2\n   =sec2A-2tanA.secA+tan2A\n  now,   put the value of secA that can b derived from the question\n             secA+tanA=2\n       or,secA=2-tanA\n                =sec2A-2tanA.2-tanA+tan2A\n           =sec2A-4tanA-2tan2A+tan2A\n             =sec2A-4tanA-tan2A\n              =tan2A-1-4tanA-tan2A\n             =-(1+4tanA)\n\nthis may b the ans(in the above solution in some places 2 means square,i hope u'll understand it)
876211
2 cos^2 x + sin^2 x = 1\n\ncos^2 x + cos^2 x + sin^2 x = 1\n\ncos^2 x + (cos^2 x + sin^2 x) = 1\n\ncos^2 x + 1 = 1\n\ncos^2 x = 0\n\ncos x = 0\n\nx = pi/2 or 3 pi/2
699510
1/tan (x+1) + 1/tan (x-1) = 1/tan (8/31)\n\nExpand tan (x+1) and tan (x-1) using the identity tan (x+y) = (tan x + tan y) / (1-tan x tan y):\n\n(1-tan x tan 1)/(tan x + tan 1) + (1+tan x tan 1)/(tan x - tan 1) = 1/tan (8/31)\n\nMultiply both sides by (tan x + tan 1) * (tan x - tan 1);\n\n(1-tan x tan 1) * (tan x - tan 1) + (1+tan x tan 1) * (tan x + tan 1) = (tan x + tan 1) * (tan x - tan 1)/tan (8/31)\n\nSimplify:\n\ntan x - tan² x tan 1 - tan 1 + tan x tan² 1 + tan x + tan² x tan 1 + tan 1 + tan x tan² 1 = (tan² x - tan² 1) * cot (8/31)\n\n2 tan x + 2 tan x tan² 1 = tan² x cot (8/31) - tan² 1 cot (8/31)\n\ncot (8/31) tan² x - (2+2 tan² 1) tan x - tan² 1 cot (8/31) = 0\n\nApplying the quadratic formula:\n\ntan x = ((2+2 tan² 1) ± √((2+2 tan² 1)² + 4 cot² (8/31) tan² 1))/(2 cot (8/31)\n\nx=arctan (((2+2 tan² 1) ± √((2+2 tan² 1)² + 4 cot² (8/31) tan² 1))/(2 cot (8/31)) ± πk\nor \nx=arctan (((2+2 tan² 1) ± √((2+2 tan² 1)² + 4 cot² (8/31) tan² 1))/(2 cot (8/31)) ± πk\n\nx≈1.216693933567969 ± πk\nx≈ -0.73097601210609 ± πk\n\nWhere k is an arbitrary integer.
837557
(secx - tanx)^2 = (1/cosx - sinx/cosx)^2 = ((1-sinx)/cosx)^2\n\n(1-sinx)/(1+sinx) \n= [(1-sinx)(1-sinx)/(1+sinx)(1-sinx)]\n\n= [(1-sinx)^2/(1-(sinx)^2)] \n= [(1-sinx)^2/(cosx)^2] \n= ((1-sinx)/cosx)^2
741518
1+(cotx)^2=(cscx)^2\n\nLeft hand side = 1 + cot²x\n= 1 + cos²x/sin²x\n= (sin²x + cos²x)/sin²x\n= 1/sin²x\n=csc²x\n= right hand side ......... QED
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.
263848
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.
1021739
Since tanx = sinx/cosx,\n\ntan(A-B) = sin(A-B)/cos(A-B)\n\nNow, write out the expression for each term on the right side of this equation.\n\nWhen you finish that, divide each term by cosAcosB and you will get the result you are seeking.
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
832720
let line 1 make an angle x with the x-axis\nso the line perpendicular toit will make an\nangle of 90*+x with the xaxis\nm1=tanx\nm2=tan(90+x)\n=-cotx\nm1m2=tanx*(-cotx)=-1\nhence proved
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
411302
cos^2 (2x) is equal to 1-sin^2(2x), remember, sin^2(x)+cos^2(x)=1
324333
Look at this link.\n\nUse the first two "sum to product" identities.\n\nOn the right you will have:\n\n= (2 / 2) (sin a / cos a) (cos b / sin b)\n\n= 1 (tan a) (cot b)\n\na = (x + y) / 2,  b = (x - y) / 2\n\nso\n\n= (tan (x+y)/2) (cot (x-y)/2)\n\nOh, darn!  You're right, the signs are reversed from this.  I wonder if the problem was given to you incorrectly.  I don't see how you could go from our result to the problem the way it was given.  Ugh!\n\nI'm sorry, but I can't go any further with this.  Good luck, and let us know if the problem was stated wrong, please.
650246
x = 8/(cos40)\ny = 8 *tan40
291468
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 !!!
714113
secant x              = 1 / cos x\nderivative             = [cos x (0) - 1 (-sin x)] / cos2 x     (I mean cos square)\n                           =  sinx / cos2 x\n                           =  tan x / cos x                (sin x / cos x = tan x)\n                           = tan x sec x                   (1 / cos x = sec x)\n\nDone! Hope this can help.