If sin^2(a)=3/4 and the terminal ray of a is in the third quadrant, then tan(a)=?
543335
sin^2(a) = 3/4 gives sin(a) = -sqrt(3)/2 or +sqrt(3)/2\n\nIf a it's in the third quadrant, sin(a) must be negative, so\nsin(a) = -sqrt(3)/2\n\nThis yields a = 210 degrees (7*pi/6 radians)\n\ntan(7pi/6) = sqrt(3)/3
730760
θ = α + β  where α = tan^-1 (1/3) and β = tan^-1 (1/2)\nie tanα = 1/3 and tanβ = 1/2\n\ntanθ = tan(α + β)\n= (tanα + tanβ)/(1 - tanαtanβ)\n= (1/3 + 1/2)/(1 - 1/3 * 1/2)\n= (5/6)/(5/6)\n= 1\nSo θ = ....., π/4, 5π/4, .....\nSo cosθ = ±1/√2 = ±√2/2
900488
cos(x) +sin(x)tan(x)=sec(x)\nmultiply by cos(x) on both sides\ncos^2(x)+sin^2(x)=1\n1=1\nI believe they are probably asking you to prove the identity
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.
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]
901871
It's 9.\ntan(beta + pi) = tan(beta + 2pi) = tan(beta) = 3\n\nSo tan beta + tan(beta + pi) + tan(beta + 2pi) = 3+3+3 = 9.\n\nYou can see why tan(beta) = tan(beta + pi) = tan(beta + 2pi) by looking at this graph of y = tan x\nhttp://www.intmath.com/TrigGrph/4_TCSC.php\nYou can see that the graph repeats itself in intervals of pi. So if you look at where y = 3, then move pi spaces to the right on the x-axis, you will see that y=3 there as well.
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
1026215
ok i got the answer to this\ni just solved it now and here's the solution.if u have time and patience then u can go through it ........\n\ntan a{tan (a+60)} + tan a {tan (a -60)} + tan (a+60)tan(a-60)\n\n= tan a(tan a + tan60) / (1-tan a tan60) + tan a (tan a - tan 60) / (1+tan a tan60) + {(tan a + tan 60) / (1- tan a tan 60) } {(tan a -tan 60)/1 + tan a tan 60)}\n\nnote ^ is to the power of,in case u didn't know.and one more thing,make sure u know all the formulae used in this problem.this will help u understand my steps better! and sqrt = sqaure root of .\n\n=( tan^2 a+ tan a sqrt3) / (1 - sqrt3tan a) + (tan ^2a - sqrt 3 tan a )/(1 + sqrt 3 tan a ) + (tan ^2a - 3)/(1 - 3 tan^2 a )--- (this particular thing (tan ^2a - 3)/(1 - 3 tan^2 a ) i got it using the formula (a+b)(a-b)=a^2-b^2 after mulipying {(tan a + tan 60) / (1- tan a tan 60) } {(tan a -tan 60)/1 + tan a tan 60)}\n\nand now that becomes\n= {(1 + sqrt3 tan a )(tan ^2 a+ sqrt 3tana) + (1 - sqrt 3 tan a)(tan^2 a - sqrt3 tan a) + tan^2 a -3} / {1 - 3tan^2a}\n=( tan^2a + sqrt3tan^3a + sqrt 3 tana + 3tan^2a + tan^2a - sqrt 3tan ^3 a - sqrt3tan a + 3tan a  + tan^2 a -3) / ((1 - 3 tan^2 a )\n\nnow u can cancel out + sqrt3tan^3a  and - sqrt3tan^3a ,+sqrt 3 tan a and -sqrt 3 tana .\n\nwhich becomes\n= (9tan^2a - 3) / (1 - 3tan^2a)\n= -3(-3tan^2a + 1) / (1 - 3tan^2 a) \n= -3\n\n\nall the best for your studies dear friend.
235956
tan(nPi/2) when n=7\n\nor tan(nPi/4) when n=2\n\nor if you need one expression,\n\n1/[Tan(nPi/2) Tan(nPi/4)].\n\nRemember, tan = sin/cos...
475675
tan x = sin x/cos x\n\nsin 2x = 2 sin x cos x\n\nso tan x sin 2x = (sin x/cos x) * (2 sin x.cos x)\n                = 2 sin^2 x\nhalf of it is = sin^2 x\n\nDifferentiate half of tan x sin 2x\n= diff of sin^2 x\n= 2 (sin x).cos x  (refer to below)\n= sin 2x    (Q.E.D.)\n\n\nDifferentiating sin^2 x.\n\nLet's call sin x as A\nand B = A^2\nthen dB/dA = diff of B * diff of A\n           = 2A * diff of A\n           = 2 sin x * cos x\n           = sin 2x
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
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
858246
Interpretation # 1\n\n2/3 tan x - 2 tan² x =1\n\nso 6 tan² x - 2 tanx + 3 = 0\n\nDiscriminant < 0 so no real solutions\n\nInterpretation # 2\n\n2 tan (x/3) - 2 tan² x =1\n\ntan 3t = tan (t + 2t)\n= (tan t + tan 2t)/(1 - tan t tan 2t)\n= (tan t + 2tan t/1 - tan² t)/( 1 - tan t*2tan t/1 - tan² t)\n= (tan t (1 - tan² t) + 2tan t)/ (1 - tan² t - 2tan² t)\n= (3tan t - tan³ t)/(1 - 3tan² t)\n\nLet tan x/3 = p Then tan x = p(3 - p²)/(1 - 3p²)\nSo 2 tan (x/3) - 2 tan² x\n= 2p - 2 (p(3 - p²)/(1 - 3p²))²\n\n= [2p(1 - 3p²)² - 2p²(3 - p²)²]/(1 - 3p²)² = 1\nSo 2p[(1 - 3p²)² - p(3 - p²)²] = (1 - 3p²)²\nie (2p - 1)(1 - 3p²)² - 2p²(3 - p²)² = 0\nSo (2p - 1)(1 - 6p² + 9p^4) - 2p²(9 - 6p² + p^4) = 0\n\nSo 18p^5 - 9p^4 - 12p³  + 6p² + 2p - 1 - 2p^6 + 12p^4 - 18p² = 0\nie 2p^6 - 18p^5 - 3p^4 + 12p³  + 12p² - 2p + 1 = 0\n\ntan(x/3) ≈ 1.09073, 9.08455 (by equation solver)\nSince 0 ≤ x < 2π then 0 ≤ x/3 < 2π/3\n\nSo x/3 ≈ 0.82877, 1.46116\nSo x ≈ 2.4863, 4.3835
882948
Just make the following substitution:\nr=sqrt(x^2+y^2)\ntheta=arctan(y/x)\nand then simplify.  Bring everything over to one side and you get your answer.\n\nIf tan(theta)=y/x, then sec(theta)=sqrt(x^2+y^2)/x.\n\nThe problem is a little ugly, but it is doable and then at the end everything ends up canceling nicely.
830426
The crux of this problem is for you to realize that the given angles are "complementary pairs", which add up to 90 degrees.\n\nYou are expected to know that the "co-function" of angle A gives the same value as the "function" of angle A's complement.\n\nSpecifically:\n\n cos(A) = sin(90-A)\n\nA graphical way of looking at this statement is that "cosine and sine are the same function, but offset by 90 degrees!"  We call that "out of phase by pi/2 or 90 degrees."\n\nThe same works for tangent and its co-function:\n\n tan(A) = cot(90-A)\n\nSo the answer to the second problem is 0.\n\nIf one of the fuctions in the first problem had been sin, the quotient would have been 1.  However, that's not the case, and I leave the upper problem as an exercise for another Answerer.\n\n Hope this helps!
820147
sin(2a)=2*sin(a)*cos(a)\ncos(2a)=cos^2(a)-sin^2(a)
728989
we have in a triangle\n\nsin B/b = sin C/c = k (say)\n\nsin B= bk  ...1\nsin C = ck  ...2 \nadditionally\n\nb^2 = c^2+a^2-2ac cos B\nso cos B = (c^2+a^2-b^2)/2ac ...3\n\ncos  C = (a^2+b^2-c^2)/2ab ...4\nfrom1 and 3 by deviding 3 by 1\ncot B = (c^2+a^2-b^2)/2abck\n\nor 2abck = (c^2+a^2-b^2) tan B  ..5\nsimilarly from 2and 4\n2abck =(a^2+b^2-c^2) tan C\n\nfrom above 2 we get the result as both are same