prove that secA/(1+sin) - cosA/(1+sinA)=tanA?
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.
940941
sin 3A = sin (2A + A)\n\n= sin 2A cos A + sinA cos 2A\n\n= (2 sin A cos A) cos A + sin A (1 - 2 sin^2A)\n\n- 2 sin A cos^2 A + sin A - 2 sin^3 A\n\n= 2 sin A (1 - sin^2 A ) + sin A - 2 sin^3 A\n\n= 2 sin A - 2 sin^3 A + sin A - 2 sin^3 A\n\n= 3 sin A - 4 sin^3 A
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
749606
L.H.S. (tanx - cotx)/(tanx + cotx) + 2cos^2x = (sinx/cosx - cosx/sinx)/(sinx/cosx + cosx/sinx) + 2cos^2x = (sin^2x - cos^2x)/(sin^2x + cos^2x) + 2cos^2x = (sin^2x - cos^2x)/1 + 2cos^2x = sin^2x - cos^2x+ 2cos^2x = sin^2x + cos^2x = 1 = R.H.S.
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
760807
Please be careful with your symbols. I believe you mean to prove\n\n    tan²x\n------------ = sin²x\n1 + tan²x\n\nwritten as tan²x / ( 1 + tan²x ) = sin²x\n\nReplace tan x with sin x / cos x to get\n\n( sin²x / cos²x ) / ( 1 +  ( sin²x / cos²x ) )\n\nMultiply numerator and denominator by cos²x to get\n\nsin²x / ( cos²x + sin²x ) = sin²x / 1 = sin²x
1048691
(r*x)*(-sin(x))+(cos(x))*(r)+\n(0)*(lnrsin(x))+(r)*(cot(x)) \n\nfor the cot(x) you end up for the derivative of lnrsin(x) is (1/sin(x)) *cos(x)=cot(x)\n\nsorry it cut off on the top. hopefully this works
820147
sin(2a)=2*sin(a)*cos(a)\ncos(2a)=cos^2(a)-sin^2(a)
868863
tan^2x+sec^2x-2tanxsecx\n=(sin^2x/cos^2x)+(1/cos^2x)-2(sinx/cosx)(1/cosx)\n=(sin^2x+1+2sinx/cos^2x\n=(1+sinx)^2/(1-sin^2x)\n=(1+sinx)/(1-sinx)
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
909136
Let I = ∫e^s sin(t - s) ds\n= e^s sin(t - s) - ∫e^s *( -1) * cos(t - s)ds\n= e^s sin(t - s) + ∫e^s cos(t - s)ds\n= e^s sin(t - s) + e^s cos(t - s) - ∫e^s * (-1) * -sin(t - s)ds\n= e^s sin(t - s) + e^s cos(t - s) - ∫e^s sin(t - s) ds\n= e^s sin(t - s) + e^s cos(t - s) - I\nSo 2I = e^s (sin(t - s) + cos(t - s)\nThus I = ½(e^s (sin(t - s) + cos(t - s)) + K\nSo ∫[s = 0 to s = t] e^s sin(t - s) ds\n= ½[e^s (sin(t - s) + cos(t - s)] from s= 0 to s = t\n= ½[(e^t (sin(t - t) + cos(t - t)) - (e^0 (sin(t - 0) + cos(t - 0))]\n= ½[(e^t(1) - 1(sint + cost)]\n= ½(e^t - sint - cost)
969937
'\nu= sec x \ndu = tan x sec x dx\ntan x dx = du/sec x = du/u\ntherefore:\nintegral of tan x dx  = integral of du/u = ln |u| + c = ln| sec x| + c
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.
874407
sec x - 1 = tan x\nsec x = tan x + 1\nsec ^2 x = tan^2 x + 2 tan x + 1\n\n[Pythagorean identity: sin^2 x + cos^2 x = 1\nDivide through by cos^2 x to get tan^2 x + 1 = sec^2 x]\n\n(tan^2 x + 1) = tan^2 x + 2 tan x + 1\n2 tan x = 0\n\ntan x = 0 where sin x = 0, which is at x = 0, 180, or 360.
798468
1=sec^2(x+y)(1+y')\nso, y'=cos^2(x+y)-1
874788
cos ^-1 0 = pi/2\n1/2 cos^-1 0 = pi/4\n\nso tan(1/2 cos^-1(0)) = tan pi/4 = 1
278456
So you know the ones for sin(2a) and cos(2a). Cool.\n\nNow, tan(2a) is the same as sin(2a)/cos(2a), since tan = sin/cos (definition of tangent).\n\nSub in the formulae for sin(2a) and cos(2a) :\n\n2sin(a)cos(a)\n----------------\ncos(a)^2-sin(a)^2\n\nDivide both numerator and denominator by (cos(a))^2\n\n2tan(a)\n----------\n1-tan(a)^2