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• 1984 AIME Problems/Problem 13
  ...y the addition formula, <math>\tan(x+y) = \frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}</math>. Let <math>a = \cot^{-1}(3)</math>, <math>b=\cot^{-1}(7)</math>, ...an(a)=\frac{1}{3},\quad\tan(b)=\frac{1}{7},\quad\tan(c)=\frac{1}{13},\quad\tan(d)=\frac{1}{21}</math>,</p></center>
  3 KB (473 words) - 12:06, 18 December 2018
• 1986 AIME Problems/Problem 15
  ...ective medians; in other words, <math>\tan \theta_1 = 1</math>, and <math>\tan \theta_2 =2</math>. ...ta_2 - \theta_1) = \frac{\tan \theta_2 - \tan \theta_1}{1 + \tan \theta_1 \tan \theta_2} = \frac{2-1}{1 + 2 \cdot 1 } = \frac{1}{3}. </cmath>
  11 KB (1,722 words) - 09:49, 13 September 2023
• 1986 AIME Problems/Problem 3
  ...tan x+\tan y=25</math> and <math>\cot x + \cot y=30</math>, what is <math>\tan(x+y)</math>? Since <math>\cot</math> is the reciprocal function of <math>\tan</math>:
  3 KB (527 words) - 10:27, 31 July 2021
• 1987 AIME Problems/Problem 15
  Let <math>\tan\angle ABC = x</math>. Now using the 1st square, <math>AC=21(1+x)</math> and ...ving, we get <math>\sin{2\theta} = \frac{1}{10}</math>. Now to find <math>\tan{\theta}</math>, we find <math>\cos{2\theta}</math> using the Pythagorean
  5 KB (838 words) - 18:05, 19 February 2022
• 1988 AIME Problems/Problem 7
  In [[triangle]] <math>ABC</math>, <math>\tan \angle CAB = 22/7</math>, and the [[altitude]] from <math>A</math> divides ...CD = 3</math>. Then <math>\tan \angle DAB = \frac{17}{h}</math> and <math>\tan \angle CAD = \frac{3}{h}</math>. Using the [[Trigonometric_identities#Angle
  1 KB (190 words) - 19:20, 27 February 2018
• 1989 AIME Problems/Problem 10
  ...\beta)^2-\tan \alpha \tan \beta}{\tan^2 \alpha + 2\tan \alpha \tan \beta +\tan^2 \beta}</math> ...sqrt{995}</math>. We see that <math>\tan \beta = \infty</math>, and <math>\tan \alpha = \sqrt{994}</math>.
  6 KB (962 words) - 14:15, 18 December 2022
• 1991 AIME Problems/Problem 15
  Let <math>a_{i} = (2i - 1) \tan{\theta_{i}}</math> for <math>1 \le i \le n</math> and <math>0 \le \theta_{i ...that that <math>S_{n} + 17 = \sum_{k = 1}^{n}(2k - 1)(\sec{\theta_{k}} + \tan{\theta_{k}})</math>.
  3 KB (497 words) - 01:43, 25 September 2020
• 1991 AIME Problems/Problem 11
  draw(Circle(((1/cos(pi/12))*cos(a*pi/12), (1/cos(pi/12))*sin(a*pi/12)), tan(pi/12))); ...h>OA</math> and <math>m \angle MOA = 15^\circ</math>. Thus <math>AM = (1) \tan{15^\circ} = 2 - \sqrt {3}</math>, which is the radius of one of the circles
  4 KB (740 words) - 19:33, 28 December 2022
• 1991 AIME Problems/Problem 9
  Suppose that <math>\sec x+\tan x=\frac{22}7</math> and that <math>\csc x+\cot x=\frac mn,</math> where <ma ...s#Pythagorean Identities|trigonometric Pythagorean identities]] <math>1 + \tan^2 x = \sec^2 x</math> and <math>1 + \cot^2 x = \csc^2 x</math>.
  10 KB (1,590 words) - 14:04, 20 January 2023
• 1993 AIME Problems/Problem 13
  Since <math>PC=100</math>, <math>PX=200</math>. So, <math>\tan(\angle OXP)=\frac{OP}{PX}=\frac{50}{200}=\frac{1}{4}</math>. Thus, <math>\tan(\angle BXA)=\tan(2\angle OXP)=\frac{2\tan(\angle OXP)}{1- \tan^2(\angle OXP)} = \frac{2\cdot \left(\frac{1}{4}\right)}{1-\left(\frac{1}{4}
  8 KB (1,243 words) - 00:26, 19 June 2022
• 1995 AIME Problems/Problem 9
  ...le sum identity gives <cmath>\tan 3x=\tan(2x+x)=\frac{3\tan x-\tan^3x}{1-3\tan^2x}.</cmath> Thus, <math>\frac{3-\tan^2x}{1-3\tan^2x}=11</math>. Solving, we get <math>\tan x= \frac 12</math>. Hence, <math>CM=\frac{11}2</math> and <math>AC= \frac{1
  7 KB (1,181 words) - 13:47, 3 February 2023
• 1996 AIME Problems/Problem 10
  Find the smallest positive integer solution to <math>\tan{19x^{\circ}}=\dfrac{\cos{96^{\circ}}+\sin{96^{\circ}}}{\cos{96^{\circ}}-\si ...2\sin{141^{\circ}}\cos{45^{\circ}}}{2\cos{141^{\circ}}\sin{45^{\circ}}} = \tan{141^{\circ}}</math>.
  4 KB (503 words) - 15:46, 3 August 2022
• 1999 AIME Problems/Problem 14
  \begin{align*}DP&=z\tan\theta\\ EP&=x\tan\theta\\
  7 KB (1,184 words) - 13:25, 22 December 2022
• 1999 AIME Problems/Problem 12
  \begin{eqnarray*} \tan \alpha & = & \frac {21}{27} \\ \tan \beta & = & \frac {21}{23} \\
  3 KB (472 words) - 15:59, 25 February 2022
• 1999 AIME Problems/Problem 11
  Given that <math>\sum_{k=1}^{35}\sin 5k=\tan \frac mn,</math> where angles are measured in degrees, and <math>m_{}</math ...ath>, we get <cmath>s = \frac{1 - \cos 175}{\sin 175} \Longrightarrow s = \tan \frac{175}{2},</cmath> and our answer is <math>\boxed{177}</math>.
  2 KB (342 words) - 18:00, 25 September 2023
• 2000 AIME I Problems/Problem 13
  ...rrow AB=\sqrt{OB^2-AO^2}=\sqrt{5^2-1.4^2}=\frac{24}{5}</math>. Then <math>\tan(\angle ABO)=\frac{OA}{AB}=\frac{7}{24}</math>, so the [[slope]] of line <ma
  3 KB (571 words) - 00:38, 13 March 2014
• 2001 AIME I Problems/Problem 5
  Note that the slope of <math>\overline{AC}</math> is <math>\tan 60^\circ = \sqrt {3}.</math> Hence, the equation of the line containing <ma
  5 KB (769 words) - 19:15, 20 July 2023
• 2003 AIME I Problems/Problem 11
  <cmath>2 > \tan 2x \Longrightarrow x < \frac 12 \arctan 2.</cmath>
  2 KB (284 words) - 13:42, 10 October 2020
• 2003 AIME I Problems/Problem 10
  pair A=(0,0), B=(2,0), C=(1,Tan(37)), M=IP(A--(2Cos(30),2Sin(30)),B--B+(-2,2Tan(23))); pair A=(0,0), B=(2,0), C=(1,Tan(37)), M=IP(A--(2Cos(30),2Sin(30)),B--B+(-2,2Tan(23)));
  7 KB (1,058 words) - 01:41, 6 December 2022
• 2003 AIME II Problems/Problem 7
  Hence <math>x=25\sin\theta=50\cos\theta</math>. Solving <math>\tan\theta=2</math>, <math>\sin\theta=\frac{2}{\sqrt{5}}, \cos\theta=\frac{1}{\s
  2 KB (323 words) - 09:56, 16 September 2022

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