secA+tanA=2  Find secA+tanA?
470919
I think you need to find A, if it is so then:\ntan A + sec A=2     square both sides\ntan^2 A + 2 tan A sec A + sec^2 A=4  since, sec^2 A= 1+tan^2 A  then:\n2 tan^2 A + 2 tan A sec A=3      since tan A=sin A/ cos A & sec A=1/cos A   then:\n2 sin^2 A/ cos^2 A + 2 sin A/ cos^2 A=3\n sin^2 A/ cos^2 A + sin A/ cos^2 A=3/2\n sin^2 A + sin A= 3/2* cos^2 A = 3/2* (1-sin^2 A)    since sin^2 A + cos^2 A =1\n5/2* sin^2 A + sin A -3/2= 0\n5 sin^2 A + 2 sin A – 3 = 0\n(5 sin A -3) (sin A +1)= 0 then,\neither sin A= 3/5 → A=36.87o \nor sin A=-1 → A=-90o (to be neglected)\ncheck: tan 36.87o=0.75 & cos 36.87o = 0.8 → sec 36.87o =1.25 then,\ntan 36.87o  + sec 36.87o =2  Q.E.D
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.
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
766844
1/tanx - 1/x=\n(x-tanx)/(xtanx)=0/0\nso use l'hopital's rule\n(1-sec^2(x))/(tanx+xsec^2(x))=\n(1-1)/(0+0)= 0/0\nso use l'hopital again\n(-2secxsecxtanx)/(sec^2(x)+\nsec^2(x)+x(2secxsecxtanx))=\n0/(2+0)=0
62277
first, distribute the 2a to the things in the parentheses:\n\n5a + (2a * 5 - 2a * b)\n\nthen, follow pemdas:\n\n5a + 10a - 2ab\n\nwhich equals:\n\n15a - 2ab
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.
650246
x = 8/(cos40)\ny = 8 *tan40
18861
tanA + tanB + tanC = tanA tanB tanC is true for any non-right angle triangle\n\nrefer the link below . adequately solved
970213
Use the Chain Rule repeatedly:\n\nOriginal function: y = (tan(sin x))^2\n\ndy/dx = 2 tan(sin x) (d/dx) tan(sin x)\n\n= 2 tan(sin x) (sec(sin x))^2 (d/dx) sin x\n\n= 2 tan(sin x) (sec(sin x))^2 cos x, QED
947972
(sec x)^2 + (tan x)^2 = 1 is not an identity.\nHere's how we arrive at that.\nStart with the identity (sin x)^2 + (cos x)^2 = 1\nDivide everything by (cos x)^2\n(sin x)^2 / (cos x)^2 + (cos x)^2 / (cos x)^2 = 1 / (cos x)^2\nYou know that (sin x) / (cos x) = tan x, \nso (sin x)^2 / (cos x)^2 = (tan x)^2\n(cos x)^2 / (cos x)^2 = 1\nand 1 / (cos x)^2 = (sec x)^2\nWe get (tan x)^2 + 1 = (sec x)^2\nWe can rearrange this to get\n(sec x)^2 - (tan x)^2 = 1\nAs you can see, this is not the same as (sec x)^2 + (tan x)^2 = 1
704759
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)
809968
(tan x- cot x)/( tan x + cot x)+ 2cos^2 x=1\n\nSenx / cos x  - cosx/senx\n---------------------------------  + 2cos^2x\nSenx / cos x  + cosx/senx\n\n(sen^2 x - cos^2x) / senx cosx\n--------------------------------------  + 2cos^2 x\n(sen^2 x + cos^2x) / senx cosx\n\nThe fraction reduces on the above:\n\nsen^2 x - cos^2 x                       + 2 cos^2 x\n\nsen^2x + cos^2x = 1\n\nproved!
652227
Taking RHS\n\ncot^2x(secx-1)\n=(secx-1)/tan^2x\n=(secx-1)/(sec^2x-1)\n=1/(secx+1)\n\nWhich is the LHS
1096282
cos^2(A) / ( 1 - sinA )\n=( 1 - sin^2(A) ) / ( 1 - sinA )\n=( 1 + sinA )( 1 - sinA ) / ( 1 - sinA )\n=1 + sinA
870571
F(x)=2[sec(x)^2] * [tan(x)] / [xcos(x)+2sin(x)]\n\nremember:\nsec(x)=1/cos(x); tan(x)=[sin(x)/cos(x)]\n\nWe rewrite the function and get:\nF(x) = [2/cos(x)^2] * [sin(x)/cos(x)] / [xcos(x)+2sin(x)] <=>\n\nF(x) = [2 /cos(x)^2] * [sin(x) /{ cos(x) * [xcos(x)+2sin(x)]} <=>\n\nF(x)= 2sin(x) / [ x*cos(x)^4 + 2sin(x)cos(x) ] =>\n\nLimF(x)= 2sin(x)/[x+2sin(x)] \nx→0\nBecause all terms with cos(x)→1 IF x→0 =>\n\nLimF(x)=2sin(x)/[x+2sin(x)] = 2sin(x)/x \nx→0\nbecause 2sin(x)→0 IF x→0 =>\n\nLim F(x)=2 because sin(x)/x →1 IF x→0 \nx→0\n=============================\nRegard from Denmark
894922
The answer is Yes!\n\n Numerator = 1 + tanx  = 1 + (sinx/cosx) = (cosx + sinx)/cosx\n Denominator = 1+ cotx = 1+ (cosx/sinx) = (sinx + cosx)/sinx\n\n Then,\n\n (1+tanx) / (1+cotx) = sinx/cosx = tanx.
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