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• 2008 AIME II Problems/Problem 5
  <cmath>\begin{align*}\tan{37}\times (1008-x) &= \tan{53} \times x\\ \frac{(1008-x)}{x} &= \frac{\tan{53}}{\tan{37}} = \frac{\sin{53}}{\cos{53}} \times\frac{\sin{37}}{\cos{37}}\end{align*
  8 KB (1,338 words) - 23:15, 28 November 2023
• 2008 AIME II Problems/Problem 11
  ...om the [[trigonometric identity|half-angle identity]], we find that <math>\tan(\theta) = \frac {3}{4}</math>. Therefore, <math>XC = \frac {64}{3}</math>. ...now drop altitude AY to solve for tan2A ; now since we know tan2A we know tan A = r/x in terms of r hence solve the resulting equation in r
  6 KB (1,065 words) - 20:12, 9 August 2022
• 2008 AIME II Problems/Problem 14
  ...tarrow (2-\sqrt{3}k)\cos x\le k\sin x\rightarrow \frac{2-\sqrt{3}k}{k}\le \tan x,</cmath>
  8 KB (1,387 words) - 11:56, 29 January 2024
• 2008 AMC 10B Problems/Problem 24
  <cmath>\cot(\theta)=\tan(5^\circ)</cmath>
  12 KB (1,944 words) - 17:15, 20 January 2024
• 1998 AHSME Problems/Problem 28
  ...c{AC(\tan 3\theta - \tan 2\theta)}{AC \tan 2\theta} = \frac{\tan 3\theta}{\tan 2\theta} - 1.</math></center> ...\tan ^2 \theta},\ \tan 3\theta = \frac{3\tan \theta - \tan^3 \theta}{1 - 3\tan^2 \theta}</math>, and
  4 KB (662 words) - 00:51, 3 October 2023
• 2000 USAMO Problems/Problem 2
  ...>. Denote <math>x=\tan{(A/2)}</math>, <math>x=\tan{(B/2)}</math>, <math>z=\tan{(C/2)}</math>, then we have, <cmath>z = \tan{(C/2)} = \tan{(90- (A+B)/2))} = \frac{1-xy}{x+y} </cmath>
  4 KB (703 words) - 18:40, 3 January 2019
• 2008 iTest Problems
  ...ath> in the interval <math>[0,2\pi)</math> that satisfy <math>\tan^2 x - 2\tan x\sin x=0</math>. Compute <math>\lfloor10S\rfloor</math>. Let a and b be the two possible values of <math>\tan\theta</math> given that <math>\sin\theta + \cos\theta = \dfrac{193}{137}</m
  71 KB (11,749 words) - 01:31, 2 November 2023
• 1989 IMO Problems/Problem 2
  ...he other triangles. Thus, the area of triangle <math>A_1BC=\frac{1}{4}a^2\tan\frac{A}{2}=\frac{1}{4}a^2\left(\frac{2r}{b+c-a}\right)</math> and similarly
  3 KB (568 words) - 11:50, 30 January 2021
• 2001 USAMO Problems/Problem 3
  ...\tan\frac{A}{2}\sin B\tan\frac{B}{2}} = 2\sqrt{\sin A\tan\frac{B}{2}\sin B\tan\frac{A}{2}} \\ &\leq \sin A\tan\frac{B}{2} + \sin B\tan\frac{A}{2} \\
  4 KB (807 words) - 10:45, 9 April 2023
• 2009 AMC 12A Problems/Problem 25
  ...ce <math>\{\theta_1, \theta_2, \theta_3...\}</math> such that <math>a_n = \tan{\theta_n}</math>, and <math>0 \leq \theta_n < 180</math>. ...+ 2}} & = \frac {\tan{\theta_n} + \tan{\theta_{n + 1}}}{1 - \tan{\theta_n}\tan{\theta_{n + 1}}} \\
  7 KB (990 words) - 07:23, 24 October 2022
• 2009 AMC 10A Problems/Problem 10
  ...get <math>xy = 3</math>, and using the fact that <math>\cot{a} = \frac{1}{\tan{a}}</math>, <math>\frac{x}{y} = 4</math>. Solving for x and y(using substit
  5 KB (879 words) - 18:57, 30 April 2024
• 2001 AMC 12 Problems/Problem 24
  ...<math>\angle ACH</math> can be simplified. Indeed, if you know that <math>\tan(75)=2+\sqrt{3}</math> or even take a minute or two to work out the sine and ...= 2 + \sqrt{3}</math>. Looking that the answer options we see that <math>\tan{75^\circ} = 2 + \sqrt{3}</math>. This means the answer is <math>D</math>.
  8 KB (1,316 words) - 22:48, 7 March 2024
• 2007 AMC 12B Problems/Problem 22
  ...}</math> and the <math>x</math>-axis is <math>30^{\circ}</math>, and <math>tan(30) = \frac{\sqrt{3}}{3}</math>.
  4 KB (707 words) - 16:36, 15 February 2021
• 2009 AMC 10B Problems/Problem 20
  ...e BAD = \angle DAC</math>. Notice <math>\tan \theta = BD</math> and <math>\tan 2 \theta = 2</math>. By the double angle identity, <cmath>2 = \frac{2 BD}{1
  2 KB (371 words) - 15:34, 15 October 2023
• 2009 AIME I Problems/Problem 12
  ...t <math>O</math> be the center of <math>\omega</math>; notice that <cmath>\tan(\angle DAO)=\dfrac{DO}{AD}=\dfrac{210/37}{144/37}=\dfrac{35}{24}</cmath> so Since we have <math>\tan OAB = \frac {35}{24}</math> and <math>\tan OBA = \frac{6}{35}</math> , we have <math>\sin {(OAB + OBA)} = \frac {1369}
  12 KB (1,970 words) - 22:53, 22 January 2024
• 2009 AIME I Problems/Problem 15
  ...BOP</math> and <math>COP</math>, with <math>BO=CO=7</math> and <math>OP=7 \tan 15=7(2-\sqrt{3})=14-7\sqrt3</math>. Then, the area of [<math>\triangle BPC<
  6 KB (1,048 words) - 19:35, 2 January 2023
• 2009 AIME II Problems/Problem 15
  <cmath>\frac{NV}{MV} = \frac{\sin (\alpha)}{\sin (90^\circ - \alpha)} = \tan (\alpha)</cmath> ...math>VW = NW + MV - 1 = \frac{1}{1+\frac{3}{4}\cot(\alpha)} + \frac{1}{1+\tan (\alpha)} - 1</math>. Taking the derivative of <math>VW</math> with respect
  11 KB (1,862 words) - 21:23, 23 May 2024
• Alternating permutation
  \sum_{n = 0}^\infty \frac{E_n}{n!} x^n = \sec x + \tan x .
  2 KB (246 words) - 12:50, 6 August 2009
• 1998 USAMO Problems/Problem 3
  ...rac {\pi}{4}\right)} + \tan{\left(a_1 - \frac {\pi}{4}\right)} + \cdots + \tan{\left(a_n - \frac {\pi}{4}\right)}\ge n - 1</cmath> Prove that <math>\tan{\left(a_0\right)}\tan{\left(a_1\right)}\cdots \tan{\left(a_n\right)}\ge n^{n + 1}</math>.
  2 KB (322 words) - 13:31, 23 August 2023
• Differentiation Rules
  If <math>y(x) = \tan x</math>, then <math>\frac{dy}{dx} = \sec^2 x</math>. Note that this follow
  2 KB (309 words) - 10:50, 4 June 2024

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