If a = 210    , Find sin 210   , cos210  , tan210   & cot 210?
924923
Ah Hi again dear \n\nwell \nSin(∏ - α) = -Sinα \nSo we have ; \nsin 210   =  sin( 180 + 30 ) = - sin30 = -1/2\ncos210   = cos( 180 + 30 ) = - cos120 =  - √3/2\ntan 210 = tan( 180 + 30 ) =√ 3/3\ncot 210  = cot ( 180 + 30 ) = √3 \n\nGood Luck dear
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
900391
csc^2(-30)+tan135+cos(-90)\n[csc(-30)]^2+tan(90+45)+cos90\n=4-1+0\n=3
247689
Use analytical trig (hint: sin function).  That should give you the answer you are looking for.
1176396
sin=opposite/hypotenuse\ncos=adjacent/hypotenuse\ntan=opposite/adjacent
308163
x can be any value.\n\ntan x = sin x / cos x\n\ncot x = cos x / sin x\n\nso tan x * cot x = sin x/cos x * cos x/sin x\n\nsin x's cross out each other\n\ncos x's cross out each other\n\nso you're left with 1.
1046178
I think...\n\n# means theta\n\nsin(90-#)*cos(90-#)=tan#/sec^2#\ncos#sin#                =(sin#/cos#)/(1/cos^2#)\ncos#sin#                =(sin#cos^2#)/cos#\ncos#sin#                =cos#sin#\n\nUsing: sin(90-#)=cos#\n           cos(90-#)=sin#\n           tan#= (sin#/cos#)\n           sec^2#=(1/cos^2#)\n\nThat work?
730612
tan x/(1 + tan x) = sin x/(sin x  + cos x)\n\ntan t = sin x/cos x\n\n(sin x/cos x)/[1 + (sin x/cos x)]\n\nMultiply big numerator and big denominator by cos x\n\nsin x/[cos x + sin x]
40284
(1) Subtitute the damn values for what you have.\n\n(2) Sin theta=opp/hyps,\n    Tan   "  =opp/adj,\n    Cos   "  =adj/hyps.\n\nNow usually a blind man follows his stick now you might know where you are at the moment - get it???
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...
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
857149
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
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.
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
1147410
sin (90 - A) = cos A so\nA = 5x - 60\n90 - A = 3x - 26\n\nAdd these equations to get\n90 = 8x - 86\n8x = 90 + 86\n8x = 176\nx = 22\n\n^_^
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 !!!
1159205
pair up the  opposite sines, for example sin(10) and sin(350) and add them:\nsin(10) + sin(350) = 0\nsin(20) + sin(340) = 0\n...\nuntil you only have sin(180) and sin(360) which both equal 0\n\nPS. I assumed you were using degrees (sexagesimal). Won't work in radians (RAD) or metric (decimal) angles.