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• 2010 AMC 12A Problems/Problem 8
  <cmath>\frac{\sqrt{3}}{2}\cos(x)=\frac{3}{2}\sin(x)\implies \tan(x)=\frac{\sqrt{3}}{3}</cmath>
  4 KB (621 words) - 15:12, 21 June 2023
• 2010 AMC 12B Problems
  ...ath> has <math>a_1=\sin x</math>, <math>a_2=\cos x</math>, and <math>a_3= \tan x</math> for some real number <math>x</math>. For what value of <math>n</ma
  12 KB (1,845 words) - 13:00, 19 February 2020
• Mock AIME 1 2010 Problems
  ...sqrt{159}}{7}</math>, and <math>ED^2 + EB^2 = 3050</math>, and that <math>\tan m \angle ACB</math> can be expressed in the form <math>\frac{a \sqrt{b}}{c}
  7 KB (1,297 words) - 01:29, 25 November 2016
• 2010 AIME I Problems/Problem 15
  <cmath>r = AX \tan(A/2) = CY \tan(C/2)</cmath> ...^2}{2\cdot 12\cdot 15} = \frac{200}{30\cdot 12}=\frac{5}{9}</math>, <math>\tan(A/2) = \sqrt{\frac{1-\cos A}{1+\cos A}} = \sqrt{\frac{9-5}{9+5}} = \frac{2}
  14 KB (2,210 words) - 13:14, 11 January 2024
• Mock AIME 2 2010 Answer Key
  Hence, <math>(4 \sin^2(x+y) - 7 \cos^2(x+y)) \leq 0</math>, yielding <math>\tan^2(x+y) = \frac{7}{4}</math>, or <math>x + y = \pm \arctan{\frac{\sqrt{7}}{2
  36 KB (6,214 words) - 20:22, 13 July 2023
• 2010 USAMO Problems/Problem 1
  <math>AY \cos(\delta)\sin(\alpha)\tan(\delta) = AY \sin(\delta)\sin(\alpha)</math>
  13 KB (2,271 words) - 14:08, 23 May 2024
• 1966 IMO Problems
  ...se sides. Prove that if <cmath> a+b=\tan{\frac{\gamma}{2}}(a\tan{\alpha}+b\tan{\beta}) </cmath> the triangle is isosceles.
  3 KB (509 words) - 12:39, 29 January 2021
• 2010 AMC 12B Problems/Problem 20
  ...ath> has <math>a_1=\sin x</math>, <math>a_2=\cos x</math>, and <math>a_3= \tan x</math> for some real number <math>x</math>. For what value of <math>n</ma ...ometric sequence, we have <math>\cos^2x=\sin x \tan x</math>. Since <math>\tan x=\frac{\sin x}{\cos x}</math>, we can rewrite this as <math>\cos^3x=\sin^2
  2 KB (375 words) - 19:52, 24 June 2022
• Mock USAJMO
  | Tan
  3 KB (295 words) - 20:01, 4 March 2020
• 2011 AMC 12A Problems/Problem 24
  ...\frac{C}{2}=90^\circ-\angle \frac{A}{2}</math>, so <math>\cot\frac{C}{2}=\tan\frac{A}{2}</math>, and we find that <math>\frac{r}{x}=\frac{x-5}{r}</math>. ...rac{B}{2}=\frac{14-x}{r}</math>, and <math>\cot\frac{D}{2}=\frac{12-x}{r}=\tan\frac{B}{2}=\frac{r}{14-x}</math>. Therefore, <math>r^2=\frac{12-x}{14-x}</m
  4 KB (753 words) - 18:58, 2 June 2022
• 2011 AMC 10B Problems/Problem 18
  ...- 12x^3 + 54x^2 - 108x + 405} \cdot \cos (\theta) = 36</cmath> and <cmath>\tan (\theta) = \frac{3}{x}</cmath> YAY!!! We have two equations for two variabl Finally, since <math>\tan (\theta) = \frac{3}{6 - 3 \sqrt{3}} = 2 + \sqrt{3}</math>, <math>\theta = \
  5 KB (782 words) - 14:29, 1 April 2024
• User:Jonjoseph
  label("$\tan \theta$", (1,.375), E);
  7 KB (964 words) - 21:43, 28 March 2022
• 2012 IMO Problems/Problem 1
  <cmath>a = R \tan \frac{B}{2} + R \tan \frac{C}{2} = R \frac{\sin \frac{B+C}{2}}{\cos \frac{B}{2} \cos \frac{C}{2}
  7 KB (1,189 words) - 01:22, 19 November 2023
• 2011 AIME I Problems/Problem 14
  Suppose the common slope of the lines is <math>m</math> and let <math>m=\tan\theta</math>. Then, we want to find <cmath>\cos 2\left(90-\theta\right)=2\c ...rt{2}\right)m^{2}&=4 \\ m^{2}&=\frac{4}{8+8\sqrt{2}} \\ \Rightarrow m^{2}=\tan^{2}\theta=\frac{\sin^{2}\theta}{\cos^{2}\theta}&=\frac{1}{2+2\sqrt{2}}.\end
  8 KB (1,344 words) - 18:39, 9 February 2023
• 2011 AIME II Problems/Problem 10
  <cmath>\textrm{tan }^{2} \alpha = \frac{\frac{9z-16}{9z}}{\frac{16}{9z}} = \frac{9z-16}{16}.< <cmath>\textrm{tan } \alpha = \textrm{tan } \angle MOP = \frac{MP}{OM} = \frac{14-y}{20}.</cmath>
  11 KB (1,720 words) - 03:12, 18 December 2023
• 2011 AIME II Problems/Problem 13
  <cmath>AP = AB \cdot (\sin 45^\circ + \cos 45^\circ \cdot \tan 30^\circ),</cmath>
  13 KB (2,055 words) - 05:25, 9 September 2022
• 1998 USAMO Problems
  ...eft(a_0-\frac{\pi}{4}\right)+ \tan \left(a_1-\frac{\pi}{4}\right)+\cdots +\tan \left(a_n-\frac{\pi}{4}\right)\geq n-1. </cmath> <cmath> \tan a_0\tan a_1 \cdots \tan a_n\geq n^{n+1}. </cmath>
  3 KB (486 words) - 06:11, 24 November 2020
• Mock Geometry AIME 2011 Problems
  ...nt to <math>OB,OC,</math> and arc <math>BC</math>. It is known that <math>\tan AOC=\frac{24}{7}</math>. The ratio <math>\frac{r_2} {r_1}</math> can be exp
  8 KB (1,349 words) - 19:10, 14 June 2022
• 2011 USAJMO Problems/Problem 3
  ...n (\theta + 120)+\tan (\theta-120)}{6}=\frac{3\tan\theta-\tan^3\theta}{2-6\tan^2\theta}</cmath> and <cmath>\begin{align*} ...{3}&=\frac{\tan\theta (\tan(\theta-120)+\tan(\theta+120))+\tan(\theta-120)\tan(\theta+120)}{12}\\
  15 KB (2,593 words) - 13:37, 29 January 2021
• LaTeX Cheat Sheet
  \sin, \cos, \tan represents <math>\sin, \cos, \tan</math>
  2 KB (318 words) - 13:47, 4 May 2024

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