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• 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 (1,231 words) - 20:06, 26 November 2023
• 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) - 19:03, 21 June 2024
• 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>.
  5 KB (816 words) - 13:39, 29 June 2024
• 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
  6 KB (1,043 words) - 10:09, 15 January 2024
• 2003 AIME I Problems/Problem 11
  <cmath>2 > \tan 2x \Longrightarrow x < \frac 12 \arctan 2.</cmath> ...math>. We can divide both sides by <math>\cos 2x</math> and we have <math>\tan 2x > 2</math>. The solutions to this occur when <math>45 \geq x > \frac{\a
  5 KB (895 words) - 15:24, 21 June 2024
• 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
• 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 (1,188 words) - 17:46, 23 June 2024
• 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

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