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• 2002 AIME II Problems/Problem 15
  ...we have that <math>\frac{y}{x}=\tan{\frac{\theta}{2}}</math>. Let <math>\tan{\frac{\theta}{2}}=m_1</math>, for convenience. Therefore if <math>(x,y)</ma <cmath>\tan{\frac{\theta}{2}}=\sqrt{\frac{1-\cos{\theta}}{1+\cos{\theta}}}</cmath>
  7 KB (1,182 words) - 09:56, 7 February 2022
• 2002 AIME II Problems/Problem 14
  We have that <math>\tan(\angle AMO)=\frac{19}{x},</math> so <cmath>\tan(\angle M)=\tan (2\cdot \angle AMO)=\frac{38x}{x^{2}-361}.</cmath>
  4 KB (658 words) - 19:15, 19 December 2021
• 2000 AIME II Problems/Problem 10
  ...</math> to get <cmath>\tan(\arctan(\tfrac{19}{r})+\arctan(\tfrac{26}{r}))+\tan(\arctan(\tfrac{37}{r})+\arctan(\tfrac{23}{r}))=n\cdot0=0.</cmath> Use the identity for <math>\tan(A+B)</math> again to get <cmath>\frac{\tfrac{45}{r}}{1-19\cdot\tfrac{26}{r^
  2 KB (399 words) - 17:37, 2 January 2024
• Law of Tangents
  <cmath> \frac{a-b}{a+b}=\frac{\tan [\frac{1}{2}(A-B)]}{\tan [\frac{1}{2}(A+B)]} . </cmath> ...2\sin s \cos d} = \frac{\tan d}{\tan s} = \frac{\tan [\frac{1}{2} (A-B)]}{\tan[ \frac{1}{2} (A+B)]} </cmath>
  2 KB (306 words) - 16:11, 21 February 2023
• University of South Carolina High School Math Contest/1993 Exam/Problem 18
  ...}{\sqrt{1 - \cos^2 (x)}} + \frac{\cos(x)}{\sqrt{1 - \sin^2 (x) }} + \frac{\tan(x)}{\sqrt{\sec^2 (x) - 1}} + \frac{\cot (x)}{\sqrt{\csc^2 (x) - 1}}</math>< | <center><math> \tan^2 x + 1 = \sec^2 x </math> </center>
  2 KB (331 words) - 00:37, 26 January 2023
• University of South Carolina High School Math Contest/1993 Exam/Problems
  ...}{\sqrt{1 - \cos^2 (x)}} + \frac{\cos(x)}{\sqrt{1 - \sin^2 (x) }} + \frac{\tan(x)}{\sqrt{\sec^2 (x) - 1}} + \frac{\cot (x)}{\sqrt{\csc^2 (x) - 1}}</cmath>
  14 KB (2,102 words) - 22:03, 26 October 2018
• Mock AIME 1 2006-2007 Problems/Problem 14
  ...f <math>AB</math>. Let <math>f(m,n)</math> denote the maximum value <math>\tan^{2}\angle AMP</math> for fixed <math>m</math> and <math>n</math> where <mat <math>\tan{\angle{OAB}} = \dfrac{OT}{AT} = \dfrac{r}{m}</math>
  3 KB (541 words) - 17:32, 22 November 2023
• Mock AIME 1 2006-2007/Problems
  ...f <math>AB</math>. Let <math>f(m,n)</math> denote the maximum value <math>\tan^{2}\angle AMP</math> for fixed <math>m</math> and <math>n</math> where <mat
  8 KB (1,355 words) - 14:54, 21 August 2020
• Mock AIME 2 2006-2007 Problems/Problem 9
  ..., <math>\frac{AY}{CY}=\sqrt 3,</math> and <math>CY=CX-BX</math>. If <math>\tan \angle APB= -\frac{a+b\sqrt{c}}{d},</math> where <math>a,b,</math> and <mat ...angle DPB)=270^\circ</math>, we have <cmath>\begin{align*}\tan\angle APB&=\tan[270^\circ-(\angle APE+\angle BPD)]\\&=\cot (\angle APE+\angle BPD)\\&=-\dfr
  2 KB (358 words) - 23:22, 3 May 2014
• Mock AIME 2 2006-2007 Problems/Problem 6
  If <math>\tan 15^\circ \tan 25^\circ \tan 35^\circ =\tan \theta</math> and <math>0^\circ \le \theta \le 180^\circ, </math> find <mat ...rc}=\frac{\sin 5^\circ(1+2\cos 20^\circ)}{\cos 5^\circ(1+2\cos 20^\circ)}=\tan 5^\circ</cmath>
  1 KB (157 words) - 10:51, 4 April 2012
• Mock AIME 2 2006-2007 Problems
  If <math>\tan 15^\circ \tan 25^\circ \tan 35^\circ =\tan \theta</math> and <math>0^\circ \le \theta \le 180^\circ, </math> find <mat ...c{BX}{CX}=\frac23</math> and <math>\frac{AY}{CY}=\sqrt 3.</math> If <math>\tan \angle APB= \frac{a+b\sqrt{c}}{d},</math> where <math>a,b,</math> and <math
  5 KB (848 words) - 23:49, 25 February 2017
• 1959 IMO Problems/Problem 4
  ...>. Thus, <math>\frac{a}{b} = \tan 15^\circ</math> and <math>\frac{a}{b} = \tan 75^\circ</math>, and so one of the angles of the triangle must be <math>15^
  6 KB (939 words) - 17:31, 15 July 2023
• Derivative/Formulas
  | <math>\frac d{dx} \tan x = \sec^2 x</math> | <math>\frac d{dx} \sec x = \sec x \tan x</math>
  3 KB (506 words) - 16:23, 11 March 2022
• Integral
  *<math>\int\tan x\,dx = \ln |\cos x| + C</math> *<math>\int \sec x\,dx = \ln |\sec x + \tan x| + C</math>
  5 KB (909 words) - 14:16, 31 May 2022
• 1970 IMO Problems/Problem 1
  & = &q \left[ \tan\left(\frac{A}{2}\right) + \tan\left(\frac{B}{2}\right) \right]\; . \end{matrix}</math> <math>\frac{r}{q} = \tan (A/2) \tan (B/2)</math>.
  2 KB (380 words) - 22:12, 19 May 2015
• 2007 iTest Problems
  ...}{4}</math> and <math>\tan{y}=\frac{1}{6}</math>, find the value of <math>\tan{x}</math>.
  33 KB (5,177 words) - 21:05, 4 February 2023
• Qq808casino
  ...opular games like baccarat, blackjack, roulette, dragon tiger, sic bo, fan tan and more. Besides, there is a selection of providers where you can expect t
  2 KB (276 words) - 03:46, 9 December 2019
• Apothem
  ...side length, <math>s</math>, the length of the apothem is <math>\frac{s}{2\tan\left(\frac{\pi}{n}\right)}</math>.
  1 KB (169 words) - 18:22, 9 March 2014
• Euler line
  ...vec BD}{\vec DA} = n = \frac {\tan \alpha – \tan \gamma}{\tan \beta – \tan \gamma} > 0.</math> ...ac {\vec AE}{\vec EC} = \frac {\tan \beta – \tan \gamma}{\tan \beta – \tan \alpha} > 0.</math>
  59 KB (10,203 words) - 04:47, 30 August 2023
• 1985 IMO Problems/Problem 1
  \begin{matrix} {CE} & = & r \tan(COE) \\
  4 KB (684 words) - 07:28, 3 October 2021

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