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• Trigonometric substitution
  ...we make use of the identity <math>\tan^2x+1=\sec^2x</math>. Set <math>x=a\tan\theta</math> and the radical will go away. However, the <math>dx</math> wil Since <math>\sec^2(\theta)-1=\tan^2(\theta)</math>, let <math>x=a\sec\theta</math>.
  1 KB (173 words) - 18:42, 30 May 2021
• Trigonometric identities
  * <math>\tan^2x + 1 = \sec^2x</math> * <math>\tan(x + y) = \frac{\tan (x) + \tan (y)}{1 - \tan (x) \tan (y)} </math>
  8 KB (1,397 words) - 21:55, 20 January 2024
• Trigonometry
  ...de opposite <math>A</math> to the side adjacent to <math>A</math>. <cmath>\tan (A) = \frac{\textrm{opposite}}{\textrm{adjacent}} = \frac{a}{b}.</cmath> ...e reciprocal of the tangent of <math>A</math>. <cmath>\cot (A) = \frac{1}{\tan (x)} = \frac{\textrm{adjacent}}{\textrm{opposite}} = \frac{b}{a}.</cmath>
  8 KB (1,217 words) - 20:15, 7 September 2023
• 2000 AMC 12 Problems
  <math>\textbf {(A)}\ \sec^2 \theta - \tan \theta \qquad \textbf {(B)}\ \frac 12 \qquad \textbf {(C)}\ \frac{\cos^2 \t
  13 KB (1,948 words) - 12:26, 1 April 2022
• 2003 AMC 12B Problems
  real r = 5/dir(54).x, h = 5 tan(54*pi/180);
  13 KB (1,987 words) - 18:53, 10 December 2022
• 2004 AMC 12B Problems
  ...of <math>\tan \angle CBE</math>, <math>\tan \angle DBE</math>, and <math>\tan \angle ABE</math> form a [[geometric progression]], and the values of <math
  13 KB (2,049 words) - 13:03, 19 February 2020
• 2006 AMC 12A Problems/Problem 19
  ...>L_2</math> and the x-axis, so <math>m=\tan{2\theta}=\frac{2\tan\theta}{1-\tan^2{\theta}}=\frac{120}{119}</math>. We also know that <math>L_1</math> and <
  2 KB (253 words) - 22:52, 29 December 2021
• 2005 AMC 12B Problems/Problem 24
  ...e positive x- axis, the answer is <math>\dfrac{\tan(BOJ) + \tan(BOJ+60) + \tan(BOJ-60)}{2}</math>. Using <math>\tan(BOJ) = 2</math>, and the tangent addition formula, this simplifies to <math
  4 KB (761 words) - 09:10, 1 August 2023
• 2005 AIME II Problems/Problem 12
  ...BG</math>). Then <math>\tan \angle EOG = \frac{x}{450}</math>, and <math>\tan \angle FOG = \frac{y}{450}</math>. ...frac{y}{450}}{1 - \frac{x}{450} \cdot \frac{y}{450}}.</cmath> Since <math>\tan 45 = 1</math>, this simplifies to <math>1 - \frac{xy}{450^2} = \frac{x + y}
  13 KB (2,080 words) - 21:20, 11 December 2022
• 2005 AIME II Problems/Problem 15
  ...y find that <math>\tan \angle OF_1T=\sqrt{69}/10</math>. Therefore, <math>\tan\angle XOT</math>, which is the desired slope, must also be <math>\sqrt{69}/ ...rac{\sqrt3\cdot\sin\theta}{2\cos\theta}=\frac65\sec\theta-\frac{\sqrt3}{2}\tan\theta</math>
  12 KB (2,000 words) - 13:17, 28 December 2020
• 2005 AIME II Problems/Problem 14
  ...5}</math>. Therefore, <math>\overline{AG} = \frac{52}{5}</math>, so <math>\tan{(\alpha)} = \frac{6}{13}</math>. Our goal now is to use tangent <math>\angl ...}</math> or <math>\frac{126}{137}</math>. Now we solve the equation <math>\tan{\angle EAG} = \frac{126}{137} = \frac{\frac{60-4x}{5}}{\frac{3x+25}{5}}</ma
  13 KB (2,129 words) - 18:56, 1 January 2024
• 2004 AIME II Problems/Problem 7
  ...B'EF=\theta</math>, so <math>\angle B'EA = \pi-2\theta</math>. Then <math>\tan(\pi-2\theta)=\frac{15}{8}</math>, or <cmath>\frac{2\tan(\theta)}{\tan^2(\theta)-1}=\frac{15}{8}</cmath> using supplementary and double angle iden
  9 KB (1,501 words) - 05:34, 30 October 2023
• 1986 AIME Problems
  ...tan x+\tan y=25</math> and <math>\cot x + \cot y=30</math>, what is <math>\tan(x+y)</math>?
  5 KB (847 words) - 15:48, 21 August 2023
• 1988 AIME Problems
  In triangle <math>ABC</math>, <math>\tan \angle CAB = 22/7</math>, and the altitude from <math>A</math> divides <mat
  6 KB (902 words) - 08:57, 19 June 2021
• 1991 AIME Problems
  Suppose that <math>\sec x+\tan x=\frac{22}7</math> and that <math>\csc x+\cot x=\frac mn,</math> where <ma draw(Circle(((1/cos(pi/12))*cos(a*pi/12), (1/cos(pi/12))*sin(a*pi/12)), tan(pi/12)));
  7 KB (1,106 words) - 22:05, 7 June 2021
• 1996 AIME Problems
  Find the smallest positive integer solution to <math>\tan{19x^{\circ}}=\dfrac{\cos{96^{\circ}}+\sin{96^{\circ}}}{\cos{96^{\circ}}-\si
  6 KB (931 words) - 17:49, 21 December 2018
• 1999 AIME Problems
  Given that <math>\sum_{k=1}^{35}\sin 5k=\tan \frac mn,</math> where angles are measured in degrees, and <math>m_{}</math
  7 KB (1,094 words) - 13:39, 16 August 2020
• 1983 AIME Problems/Problem 4
  ...an{(\angle OAB)}=\frac{2-\frac{1}{3}}{1+2\cdot \frac{1}{3}}</cmath><cmath>\tan{(\angle OAB)}=1</cmath><cmath>\cos{(\angle OAB)}=\frac{1}{\sqrt{2}}</cmath>
  11 KB (1,741 words) - 22:40, 23 November 2023
• 1983 AIME Problems/Problem 15
  .../math>, we have <math>OM = \sqrt{OB^2 - BM^2} =4</math>. This gives <math>\tan \angle BOM = \frac{BM}{OM} = \frac 3 4</math>. ...efore, since <math>\angle AOM</math> is clearly acute, we see that <cmath>\tan \angle AOM =\frac{\sqrt{1 - \cos^2 \angle AOM}}{\cos \angle AOM} = \frac{\s
  19 KB (3,221 words) - 01:05, 7 February 2023
• 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 (545 words) - 23:44, 12 October 2023
• 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 ...lpha + \tan \beta}{1 - \tan \alpha \cdot \tan \beta}</math>, we get <math>\tan (\alpha + \beta) = \dfrac {\frac {20}{h}}{\frac {h^2 - 51}{h^2}} = \frac {2
  1 KB (178 words) - 23:25, 20 November 2023
• 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>.
  8 KB (1,401 words) - 21:41, 20 January 2024
• 1990 AIME Problems/Problem 7
  ...th>-axis and <math>PR</math>. The equation of the angle bisector is <math>\tan\left(\frac{\alpha+\beta}{2}\right)</math>. ...}}{\frac{18}{25}}}=\pm\sqrt{\frac{16}{9}}=\pm\frac{4}{3}</math> and <math>\tan\left(\frac{\beta}{2}\right)=\pm\sqrt{\frac{1-\cos(\beta)}{1+\cos(\beta)}}=\
  8 KB (1,319 words) - 11:34, 22 November 2023
• 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>.
  4 KB (658 words) - 16:58, 10 November 2023
• 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 (1,231 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) - 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>.
  4 KB (614 words) - 04:38, 8 December 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
  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>
  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
• 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
• Asymptote (geometry)
  ...the vertical asymptotes of 1) <math>y = \frac{1}{x^2-5x}</math> 2) <math>\tan 3x</math>. 2) Since <math>\tan 3x = \frac{\sin 3x}{\cos 3x}</math>, we need to find where <math>\cos 3x =
  4 KB (664 words) - 11:44, 8 May 2020
• Mock AIME 3 Pre 2005 Problems
  The value of <math>\tan\left(\Omega\right)</math> can be expressed as <math>\frac{m}{n}</math>, whe
  7 KB (1,135 words) - 23:53, 24 March 2019
• Mock AIME 3 Pre 2005 Problems/Problem 15
  The value of <math>\tan\left(\Omega\right)</math> can be expressed as <math>\frac{m}{n}</math>, whe ...ric substitution; namely, define <math>\theta</math> such that <math> x = \tan{\theta}</math>. Then the RHS becomes
  2 KB (312 words) - 10:38, 4 April 2012
• 1960 IMO Problems/Problem 3
  \tan{\alpha}=\frac{4nh}{(n^2-1)a}. ...c}{b}\cdot\frac{n-1}{n+1}</math>, and <math>\text{slope}</math><math>(QA)=\tan{\angle QAB}=\frac{c}{b}\cdot\frac{n+1}{n-1}</math>.
  3 KB (501 words) - 00:14, 17 May 2015
• 1960 IMO Problems
  \tan{\alpha}=\frac{4nh}{(n^2-1)a}.
  3 KB (511 words) - 21:21, 20 August 2020
• 2007 AIME I Problems/Problem 9
  ...{1 - \cos \theta}{1 + \cos \theta}}</math>). We see that <math>\frac rx = \tan \frac{180 - \theta}{2} = \sqrt{\frac{1 - \cos (180 - \theta)}{1 + \cos (180 ...We see that <math>\frac rx = \tan \left(\frac{180 - 2\theta}{2}\right) = \tan (90 - \theta)</math>. In terms of <math>r</math>, we find that <math>x = \f
  11 KB (1,851 words) - 12:31, 21 December 2021
• 2007 AIME I Problems/Problem 12
  ...th>[\triangle EFB'] = \frac{1}{2} (FB' \cdot EF) = \frac{1}{2} (FB') (FB' \tan 75^{\circ})</math>. With some horrendous [[algebra]], we can calculate [\triangle EFB'] &= \frac{1}{2}\tan (30 + 45) \cdot (20 - 5\sqrt{2} - 5\sqrt{6})^2 \\
  10 KB (1,458 words) - 20:50, 3 November 2023
• 2007 AIME II Problems/Problem 15
  ...2}{3}</math> according to half angle formula. Similarly, we can find <math>tan\angle NCK=\frac{1}{2}</math>. So we can see that <math>JK=ON=14-\frac{7x}{2
  11 KB (2,099 words) - 17:51, 4 January 2024
• 1961 IMO Problems/Problem 5
  <math>b \tan{\frac{\omega}{2}} \le c < b</math> ...we require <math>AX \geqslant AC > AB</math>. But <math>\frac{AB}{AX} = \tan{\frac{\omega}{2}}</math>, so we get the condition in the question
  1 KB (205 words) - 04:12, 7 June 2021
• LaTeX:Commands
  ...||\cos||<math>\textstyle \sin</math>||\sin||<math>\textstyle \tan</math>||\tan
  12 KB (1,898 words) - 15:31, 22 February 2024
• 2006 AIME II Problems/Problem 6
  ...Also note that <cmath>AB = 1 = \overline{AA'} + \overline{A'B} = \frac{x}{\tan(15)} + x</cmath> Using the fact <math>\tan(15) = 2-\sqrt{3}</math>, this yields <cmath>x = \frac{1}{3+\sqrt{3}} = \fra
  7 KB (1,067 words) - 12:23, 8 April 2024
• 1961 IMO Problems
  <math>b \tan{\frac{\omega}{2}} \le c < b</math>
  3 KB (425 words) - 21:18, 20 August 2020
• 2004 AMC 12A Problems/Problem 18
  ...e CGF</math>. Thus, <math>\angle EGF=\angle FCG</math> and <math>\tan EGF=\tan FCG=\frac{1}{2}</math>. Solving <math>EF=\frac{1}{2}</math>. Adding, the an
  5 KB (738 words) - 13:11, 27 March 2023
• 2004 AMC 10A Problems/Problem 20
  E = (0,Tan(15)); F = (1 - Tan(15),1);
  4 KB (710 words) - 02:47, 18 April 2024
• 1999 AHSME Problems
  ...al number such that <math>\sec x - \tan x = 2</math>. Then <math>\sec x + \tan x =</math>
  13 KB (1,945 words) - 18:28, 19 June 2023
• 1960 IMO Problems/Problem 6
  <cmath>\tan\left(\frac{\theta}{2}\right) = \frac{1}{x} = \frac{\sqrt{2}}{4}</cmath> ...3</math> and <math>V_1 = \frac{\pi a^2 \times H_1 H_2}{3} = \frac{\pi a^3 \tan (\angle A_1 A H_1) }{3}</math> .
  7 KB (1,214 words) - 18:49, 29 January 2018
• 1992 USAMO Problems/Problem 2
  Consider the points <math>M_k = (1, \tan k^\circ)</math> in the coordinate plane with origin <math>O=(0,0)</math>, f ...hen the left hand side of the equation simplifies to <math>\tan 89-\tan 0=\tan 89=\frac{\sin 89}{\cos 89}=\frac{\cos 1}{\sin 1}</math> as desired.
  4 KB (628 words) - 07:41, 19 July 2016
• 2000 AMC 12 Problems/Problem 17
  <math>\text {(A)}\ \sec^2 \theta - \tan \theta \qquad \text {(B)}\ \frac 12 \qquad \text {(C)}\ \frac{\cos^2 \theta ...<cmath> \frac{OB}{OC} = \frac{AB}{AC} \Longrightarrow AC \sec \theta = OC \tan \theta </cmath> We multiply both sides by <math>\cos \theta</math> to simpl
  6 KB (979 words) - 12:50, 17 July 2022
• Construction
  21. Construct <math>sin C, cos C, tan C</math> given unit segment <math>1</math> and acute angle <math>C</math>.
  3 KB (443 words) - 20:52, 28 August 2014
• 1995 USAMO Problems
  ..., </math> <math>\tan, \; \sin^{-1}, \; \cos^{-1}, \,</math> and <math>\, \tan^{-1} \,</math> buttons. The display initially shows 0. Given any positive
  3 KB (540 words) - 13:31, 4 July 2013
• 2004 AMC 12B Problems/Problem 24
  ...of <math>\tan \angle CBE</math>, <math>\tan \angle DBE</math>, and <math>\tan \angle ABE</math> form a [[geometric progression]], and the values of <math ...a)\tan(DBE + \alpha) = \frac {\tan^2 DBE - \tan^2 \alpha}{1 - \tan ^2 DBE \tan^2 \alpha},
  2 KB (302 words) - 19:59, 3 July 2013
• 1995 USAMO Problems/Problem 2
  ..., </math> <math>\tan, \; \sin^{-1}, \; \cos^{-1}, \,</math> and <math>\, \tan^{-1} \,</math> buttons. The display initially shows 0. Given any positive <cmath> \cos \tan^{-1} \sqrt{(n-m)/m} = \sqrt{m/n} . </cmath>
  3 KB (516 words) - 00:18, 6 April 2020
• 2008 AMC 12A Problems
  ...midpoint of <math>BC</math>. What is the largest possible value of <math>\tan{\angle BAD}</math>?
  13 KB (2,025 words) - 13:56, 2 February 2021
• 2008 AMC 12A Problems/Problem 24
  ...dpoint]] of <math>BC</math>. What is the largest possible value of <math>\tan{\angle BAD}</math>? ..., and since <math>\tan\angle BAF = \frac{2\sqrt{3}}{x-2}</math> and <math>\tan\angle DAE = \frac{\sqrt{3}}{x-1}</math>, we have
  3 KB (513 words) - 14:35, 7 June 2018
• 2008 AMC 12B Problems/Problem 24
  ...a</math> with the x-axis and passes through the origin has equation <math>\tan(\theta)x</math>, so the line through <math>A_0</math> and <math>B_1</math>
  9 KB (1,482 words) - 13:52, 4 April 2024
• 2008 AIME I Problems/Problem 8
  Since we are dealing with acute angles, <math>\tan(\arctan{a}) = a</math>. Note that <math>\tan(\arctan{a} + \arctan{b}) = \dfrac{a + b}{1 - ab}</math>, by tangent additio
  3 KB (490 words) - 22:36, 28 November 2023
• 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
• 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 (1,849 words) - 19:43, 2 January 2023
• 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 (288 words) - 00:53, 26 March 2018
• 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 (2,178 words) - 14:14, 11 September 2021
• 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
• 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 (315 words) - 21:13, 28 February 2022
• 1999 AHSME Problems/Problem 15
  ...number such that <math> \sec x - \tan x = 2</math>. Then <math> \sec x + \tan x =</math> ...(\sec x + \tan x) = \sec^{2} x - \tan^{2} x = 1</math>, so <math>\sec x + \tan x = \boxed{\textbf{(E)}\ 0.5}</math>.
  931 bytes (144 words) - 19:36, 1 May 2023
• 1984 AHSME Problems
  ...}) \qquad \mathrm{(D) \ }\tan{15^\circ} \qquad \mathrm{(E) \ } \frac{1}{4}\tan{60^\circ} </math>
  13 KB (1,879 words) - 14:00, 19 February 2020
• Trigonometric Equations
  <math>sin^2(x) + tan^2(x) = -cos^2(x) + \frac{1}{sin^2(x) + cos^2(x)}</math> <math>sin^2(x) + cos^2(x) + tan^2(x) = \frac{1}{sin^2(x) + cos^2(x)}</math>
  8 KB (1,351 words) - 20:30, 10 July 2016
• 1984 AHSME Problems/Problem 29
  The slope we are looking for is equivalent to <math>\tan (\theta + 45)</math> where <math>\angle AOX = \theta</math>. Using tangent <cmath> \tan (\theta + 45)= \frac{\tan \theta + \tan 45}{1-\tan\theta\tan 45} = \frac{\frac{1}{\sqrt{2}}+1}{1-\frac{1}{\sqrt{2}}}=3+2\sqrt{2}</cmath>
  4 KB (614 words) - 20:09, 12 September 2022
• 1984 AHSME Problems/Problem 15
  ...times\sin{3x}}{\cos{2x}\times\cos{3x}}=1 </math>, or <math> \tan{2x}\times\tan{3x}=1 </math>. ...identity]] <math> -\tan{x}=\tan{-x} </math>, we have <math> \tan{2x}\times\tan{-3x}=-1 </math>.
  3 KB (493 words) - 18:16, 4 June 2021
• 1966 IMO Problems/Problem 2
  <cmath> a+b=\tan{\frac{\gamma}{2}}(a\tan{\alpha}+b\tan{\beta}) </cmath> ...at as <math>\gamma = \pi -\alpha-\beta</math> then and the identity <math>\tan\left(\frac \pi 2 - x \right)=\cot x</math> our equation becomes:
  2 KB (416 words) - 17:54, 13 January 2022
• 1984 AHSME Problems/Problem 23
  ...}) \qquad \mathrm{(D) \ }\tan{15^\circ} \qquad \mathrm{(E) \ } \frac{1}{4}\tan{60^\circ} </math> ...irc}}{2\cos{15^\circ}\cos{5^\circ}}=\frac{\sin{15^\circ}}{\cos{15^\circ}}=\tan{15^\circ}, \boxed{\text{D}} </math>.
  1 KB (159 words) - 12:52, 5 July 2013
• 1984 AHSME Problems/Problem 26
  ...can also use trig manipulation on <math>BCE</math> to get that <math>CE=a\tan{\beta}</math>. <math>[BED]=\frac{BD\cdot CE}{2}=\frac{ac\cos{\beta}\tan{\beta}}{4}=\frac{ac\sin{\beta}}{4}</math>
  2 KB (303 words) - 20:28, 2 October 2023
• 1985 AHSME Problems
  ...= 25 \degree</math>, then the value of <math>\left(1+\tan A\right)\left(1+\tan B\right)</math> is ...\qquad \mathrm{(C) \ } 1+\sqrt{2} \qquad \mathrm{(D) \ } 2\left(\tan A+\tan B\right) \qquad \mathrm{(E) \ }\text{none of these} </math>
  17 KB (2,488 words) - 03:26, 20 March 2024
• 1985 AHSME Problems/Problem 16
  ...= 25 \degree</math>, then the value of <math>\left(1+\tan A\right)\left(1+\tan B\right)</math> is ...\qquad \mathrm{(C) \ } 1+\sqrt{2} \qquad \mathrm{(D) \ } 2\left(\tan A+\tan B\right) \qquad \mathrm{(E) \ }\text{none of these} </math>
  5 KB (904 words) - 22:25, 19 March 2024
• 2012 AMC 12B Problems/Problem 25
  ...counter-clockwise order and right angle at <math>A</math>, let <math>f(t)=\tan(\angle{CBA})</math>. What is <cmath>\prod_{t\in T} f(t)?</cmath> ...on <math>A'B'C'</math> labeled that way will give us <math>\tan CBA \cdot \tan C'B'A' = 1</math>. First we consider the reflection about the line <math>y=
  2 KB (356 words) - 17:10, 4 April 2020
• Mock Geometry AIME 2011 Problems/Problem 1
  Now we have <math> BE=BA\cdot\tan\angle EAB=1\cdot\tan30^\circ=\frac{\sqrt{3}}{3} </math>. Finally, <math> [A
  2 KB (376 words) - 22:06, 23 December 2022
• Mock Geometry AIME 2011 Problems/Problem 11
  ...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 ...h>, so <math> \cos AOC=\frac{7}{25} </math>. Now, we have <math> \tan FOD=\tan\frac{AOC}{2}=\sqrt{\frac{1-\cos AOC}{1+\cos AOC}}=\sqrt{\frac{1-\frac{7}{25
  3 KB (432 words) - 14:12, 2 January 2012
• Mock Geometry AIME 2011 Problems/Problem 8
  ...t triangles <math>\Delta PO_1H, \Delta O_1S_1O_2</math>. Similarly, <math>\tan{\angle PO_2O_1}=\frac{h}{b}=\frac{52}{39}</math> using right triangles <mat
  3 KB (522 words) - 21:25, 3 January 2012
• 1989 AHSME Problems
  <math> \cot 10+\tan 5 = </math> Find the sum of the roots of <math>\tan^2x-9\tan x+1=0</math> that are between <math>x=0</math> and <math>x=2\pi</math> radi
  15 KB (2,247 words) - 13:44, 19 February 2020
• 1991 AHSME Problems/Problem 21
  \text{(C) } tan^2\theta\quad
  738 bytes (126 words) - 21:56, 17 October 2016
• 1989 AHSME Problems/Problem 28
  Find the sum of the roots of <math>\tan^2x-9\tan x+1=0</math> that are between <math>x=0</math> and <math>x=2\pi</math> radi ...</math> are positive and distinct, so by considering the graph of <math>y=\tan x</math>, the smallest two roots of the original equation <math>x_1,\ x_2</
  2 KB (282 words) - 21:14, 2 March 2019
• 1989 AHSME Problems/Problem 14
  <math>\cot 10+\tan 5=</math> We have <cmath>\cot 10 +\tan 5=\frac{\cos 10}{\sin 10}+\frac{\sin 5}{\cos 5}=\frac{\cos10\cos5+\sin10\si
  534 bytes (69 words) - 16:11, 25 February 2022
• 2012 AMC 12B Problems
  ...counter-clockwise order and right angle at <math>A</math>, let <math>f(t)=\tan(\angle{CBA})</math>. What is <cmath>\prod_{t\in T} f(t)?</cmath>
  20 KB (2,681 words) - 09:47, 29 June 2023
• 2012 AMC 12B Problems/Problem 17
  ...</math> so that <math>\cos\theta=\sin(90-\theta)=s/6</math>. Then <math>m=\tan\theta=3</math>. Substituting into <math>\left(\frac{4m^2+10}{m^2+1},\frac{6 ...</math>. Setting <math>6\sin\theta=-2\cos\theta</math>, we get that <math>\tan\theta=-1/3</math>. This means <math>-1/3</math> is the slope of line <math>
  12 KB (2,183 words) - 21:05, 23 December 2023
• 2012 AIME I Problems
  ...\angle C.</math> If <math>\frac{DE}{BE} = \frac{8}{15},</math> then <math>\tan B</math> can be written as <math>\frac{m \sqrt{p}}{n},</math> where <math>m
  10 KB (1,617 words) - 14:49, 2 June 2023
• 2012 AIME I Problems/Problem 12
  ...\angle C.</math> If <math>\frac{DE}{BE} = \frac{8}{15},</math> then <math>\tan B</math> can be written as <math>\frac{m \sqrt{p}}{n},</math> where <math>m ...<math>FD = 4a\sqrt{3}</math>, and <math>FB = 11a</math>. Finally, <math>\tan{B} = \tfrac{DF}{FB}=\tfrac{4\sqrt{3}a}{11a} = \tfrac{4\sqrt{3}}{11}</math>.
  9 KB (1,523 words) - 12:23, 7 September 2022
• Mock AIME II 2012 Problems/Problem 6
  pair tan = reflect(origin,(5,y))*(0,y); draw((5,y)--tan,linetype("4 4"));
  3 KB (478 words) - 03:06, 5 April 2012
• 2006 SMT/Calculus Problems
  Evaluate: <math> \int(x\tan^{-1}x)dx </math> <cmath> \frac{d}{dx}\tanh^{-1}\tan x </cmath>
  3 KB (525 words) - 13:59, 27 May 2012
• 2006 SMT/Team Problems
  ...exist two positive numbers <math> x </math> such that <math> \sin(\arccos(\tan(\arcsin(x))))=x </math>. Find the product of the two possible <math> x </ma
  6 KB (910 words) - 17:32, 27 May 2012
• 1953 AHSME Problems/Problem 50
  ...rac{2}{3}</math>, <math>\tan{\frac{B}{2}} = \frac{1}{2}</math>, and <math>\tan{\frac{C}{2}} = \frac{4}{x},</math> so we have the equation <math>\frac{1}{2
  2 KB (314 words) - 21:17, 31 December 2023
• 2013 AMC 12B Problems
  ..., Q=(\cot x, \cot^2 x), R=(\sin x, \sin^2 x)</math> and <math>S =(\tan x, \tan^2 x)</math> are the vertices of a trapezoid. What is <math>\sin(2x)</math>?
  16 KB (2,459 words) - 02:46, 30 January 2021
• 2013 AMC 12B Problems/Problem 20
  ..., Q=(\cot x, \cot^2 x), R=(\sin x, \sin^2 x)</math> and <math>S =(\tan x, \tan^2 x)</math> are the vertices of a trapezoid. What is <math>\sin(2x)</math>? Let <math>f,g,h,j</math> be <math>\sin, \cos, \tan, \cot</math> (not respectively). Then we have four points <math>(f,f^2),(g,
  2 KB (375 words) - 00:54, 28 September 2021
• 1982 USAMO Problems/Problem 3
  ...} \cdot \frac{r}{s-b} \cdot \frac{r}{s-c} = \frac{1}{4} \tan A/2 \tan B/2 \tan C/2.</cmath> Lemma. <math>\tan x \tan (A - x)</math> is increasing on <math>0 < x < \frac{A}{2}</math>, where <ma
  2 KB (376 words) - 23:29, 18 May 2015
• 2013 AIME I Problems/Problem 14
  ...tter. Only the numerator, because we are trying to find <math>\frac{P}{Q}=\tan\text{arg}(\Sigma)</math> a PROPORTION of values. So denominators would canc
  10 KB (1,641 words) - 20:03, 3 January 2024
• 1991 AHSME Problems
  \textbf{(C) } \tan^2\theta\qquad
  16 KB (2,451 words) - 04:27, 6 September 2021
• 2003 AMC 12B Problems/Problem 10
  real r = 5/dir(54).x, h = 5 tan(54*pi/180);
  1 KB (237 words) - 23:06, 3 February 2020
• Hyperbolic trig functions
  <math>\tanh(x)= -1\tan{iz}</math>
  423 bytes (78 words) - 23:33, 22 May 2013
• 2013 USAJMO Problems/Problem 5
  <cmath>\frac{XC}{CY}=\tan {\angle CYZ}=\tan (90-\alpha)</cmath> <cmath>\frac{CQ}{CY}=\tan {\angle CYQ}=\tan (\alpha+\beta).</cmath>
  7 KB (1,250 words) - 18:05, 1 October 2021
• 2014 AIME II Problems/Problem 12
  <math>\tan{\frac{3A}{2}}\tan{\frac{3B}{2}}=1</math> Note that <math>\tan{x}=\frac{1}{\tan(90-x)}</math>, or <math>\tan{x}\tan(90-x)=1</math>
  5 KB (875 words) - 17:56, 2 October 2023
• 2013 IMO Problems/Problem 4
  ...an{\angle{C}}-1)m}{i\tan{\angle{C}}}.</cmath> We wish to simplify <math>(i\tan{\angle{C}}-1)m</math> first. Note that <cmath>m=\frac{|CM|}{|CA|}\cdot(a)=\ (i\tan{\angle{C}}-1)m&=(i\tan{\angle{C}}-1)((|BC|)\cos{\angle{C}}(\cos{\angle{C}}+i\sin{\angle{C}}))\\
  11 KB (1,991 words) - 01:31, 19 November 2023
• 2014 AMC 10A Problems/Problem 22
  ...ow EC=20-10 \sqrt 3</math>. (It is important to memorize the sin, cos, and tan values of <math>15^\circ</math> and <math>75^\circ</math>.) Therefore, we h
  12 KB (1,821 words) - 18:16, 29 October 2023
• 2014 AMC 12A Problems/Problem 25
  ...an angle <math>\theta</math> relative to the coordinate axis, where <math>\tan\theta = \tfrac 34</math>. We rotate the coordinate axis by angle <math>\the
  4 KB (661 words) - 16:18, 2 September 2022
• 2014 AMC 12B Problems/Problem 19
  ...tan {\theta})^4 - 3(\tan {\theta})^2 + 1 = 0.</math> This gives us <math>(\tan {\theta})^2 = \dfrac{3+\sqrt{5}}{2}\longrightarrow \boxed{E}</math>
  4 KB (703 words) - 16:24, 9 September 2022
• 2014 AIME I Problems
  ...\left(\tfrac{w-z}{z}\right) </math>. The maximum possible value of <math>\tan^2 \theta</math> can be written as <math>\tfrac{p}{q}</math>, where <math>p<
  9 KB (1,472 words) - 13:59, 30 November 2021
• 2014 AIME I Problems/Problem 7
  ...\left(\tfrac{w-z}{z}\right) </math>. The maximum possible value of <math>\tan^2 \theta</math> can be written as <math>\tfrac{p}{q}</math>, where <math>p< We know that <math>\tan{\theta}</math> is equal to the imaginary part of the above expression divid
  5 KB (782 words) - 20:25, 10 October 2023
• 1973 AHSME Problems
  ...nd <math> \sin \frac12 \theta = \sqrt{\frac{x-1}{2x}}</math>, then <math> \tan \theta</math> equals
  18 KB (2,788 words) - 13:55, 20 February 2020
• 1980 AHSME Problems/Problem 12
  ..._2)}{1-\tan^2(\theta_2)} = 4\tan(\theta_2)</math>. Solving, we have <math>\tan(\theta_2) = 0, \dfrac{\sqrt{2}}{2}</math>. But line <math>L_1</math> is not
  2 KB (237 words) - 18:02, 20 March 2018
• 1987 AHSME Problems
  ...c})+\log_{10}(\tan 3^{\circ})+\cdots+\log_{10}(\tan 88^{\circ})+\log_{10}(\tan 89^{\circ}). </math>
  16 KB (2,291 words) - 13:45, 19 February 2020
• 1986 AHSME Problems
  <math>\textbf{(A)}\ \tan \theta = \theta\qquad \textbf{(B)}\ \tan \theta = 2\theta\qquad
  17 KB (2,512 words) - 18:30, 12 October 2023
• 1983 AHSME Problems
  ...angle at <math>C</math>. If <math>\sin A = \frac{2}{3}</math>, then <math>\tan B</math> is If <math>\tan{\alpha}</math> and <math>\tan{\beta}</math> are the roots of <math>x^2 - px + q = 0</math>, and <math>\co
  15 KB (2,309 words) - 23:43, 2 December 2021
• 1978 AHSME Problems
  If <math>\sin x+\cos x=1/5</math> and <math>0\le x<\pi</math>, then <math>\tan x</math> is
  15 KB (2,432 words) - 01:06, 22 February 2024
• 1972 AHSME Problems
  If <math>\tan x=\dfrac{2ab}{a^2-b^2}</math> where <math>a>b>0</math> and <math>0^\circ <x real x = 6-h*tan(t);
  17 KB (2,732 words) - 13:54, 20 February 2020
• 2007 iTest Problems/Problem 17
  ...}{4}</math> and <math>\tan{y}=\frac{1}{6}</math>, find the value of <math>\tan{x}</math>. <math>\tan(x+\arctan\frac{1}{6})=\tan\frac{\pi}{4}=1</math>
  2 KB (266 words) - 21:30, 4 February 2023
• 2007 IMO Problems/Problem 4
  ...PK = \dfrac{1}{2}a \tan \dfrac{1}{2}C</math> and <math>QL = \dfrac{1}{2}b \tan \dfrac{1}{2}C</math> from right triangles <math>\triangle PKC</math> and <m <cmath>= \dfrac{\frac{1}{2}a\tan\frac{1}{2}C \cdot (a + b)}{a\sin\frac{1}{2}C} </cmath>
  8 KB (1,480 words) - 14:52, 5 August 2022
• 1988 AHSME Problems/Problem 13
  ...ation by <math>\cos{(x)}</math> to get <cmath>\frac{\sin{(x)}}{\cos{(x)}}=\tan{(x)}=3.</cmath>
  2 KB (402 words) - 20:53, 24 August 2021
• 1987 AHSME Problems/Problem 20
  ...c})+\log_{10}(\tan 3^{\circ})+\cdots+\log_{10}(\tan 88^{\circ})+\log_{10}(\tan 89^{\circ}). </math> ...b}</math>, the answer is <math>\log_{10} {\tan 1^\circ \tan 2^\circ \dots \tan 89^\circ} = \log_{10} 1 = 0.</math> <math>\boxed{\textbf{(A)}}.</math>
  1 KB (164 words) - 12:42, 31 March 2018
• 1986 AHSME Problems/Problem 21
  <math>\textbf{(A)}\ \tan \theta = \theta\qquad \textbf{(B)}\ \tan \theta = 2\theta\qquad
  2 KB (301 words) - 18:50, 1 April 2018
• 1986 AHSME Problems/Problem 28
  ...}{DP}, DP = \frac{1}{\tan 54}</cmath>Therefore, <math>AB = 2DP = \frac{2}{\tan 54}</math>. ...efore, <math>AO + AQ + AR = AO + 2AQ = \frac{1}{\sin 54}+\frac{4 \sin 72}{\tan 54} = \frac{1}{\sin 54} + 8 \sin 36 \cos 54 = \frac{1}{\cos 36} + 8-8\cos^2
  4 KB (702 words) - 17:13, 17 April 2020
• Angle Addition Formulas (Trigonometry)
  ..., we get<cmath>\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}.</cmath>
  2 KB (458 words) - 19:24, 2 February 2020
• 2015 AMC 10A Problems/Problem 19
  ...> is an isosceles <math>30 - 75 - 75</math> triangle. Thus, <math>DF = CF \tan 15^\circ = \dfrac{5}{\sqrt{2}} (2 - \sqrt{3})</math> by the Half-Angle form
  9 KB (1,513 words) - 19:38, 12 November 2023
• 2015 AIME II Problems/Problem 15
  Setting the two areas equal, we get <cmath>\tan A = \frac{\sin A}{\cos A} = 8 \iff \sin A = \frac{8}{\sqrt{65}}, \cos A = \ <cmath>\tan{\angle{CYE}} = \frac{1}{8}</cmath>
  31 KB (5,086 words) - 19:15, 20 December 2023
• 2015 USAMO Problems/Problem 2
  Let angle <math>\angle XAB=A</math>, which is an acute angle, <math>\tan{A}=t</math>, then <math>X=(1-a,at)</math>.
  5 KB (902 words) - 09:58, 20 August 2021
• 2015 USAJMO Problems/Problem 3
  Let angle <math>\angle XAB=A</math>, which is an acute angle, <math>\tan{A}=t</math>, then <math>X=(1-a,at)</math>.
  4 KB (760 words) - 16:45, 29 April 2020
• 2016 AMC 12B Problems/Problem 13
  From Alice's point of view, <math>\tan(\theta)=\frac{z}{y}</math>. <math>\tan{30}=\frac{\sin{30}}{\cos{30}}=\frac{1}{\sqrt{3}}</math>. So, <math>y=z*\sqr From Bob's point of view, <math>\tan(\theta)=\frac{z}{x}</math>. <math>\tan{60}=\frac{\sin{60}}{\cos{60}}=\sqrt{3}</math>. So, <math>x = \frac{z}{\sqrt
  6 KB (803 words) - 00:37, 2 November 2023
• 2016 AIME I Problems/Problem 9
  <cmath>\frac{31}{40} \geq \tan\theta</cmath> However, <math>\tan\theta = \tan(\frac{90-A}{2}) = \frac{\sin(90-A)}{\cos(90-A)+1} = \frac{\cos A}{\sin A +
  9 KB (1,526 words) - 02:31, 29 December 2021
• 2016 AIME II Problems/Problem 14
  ...and <math>KQ=y</math>, assuming WLOG <math>x>y</math>, we must have <math>\tan(120^{\circ})=-\sqrt{3}=\dfrac{\dfrac{x+y}{100 \sqrt{3}}}{\dfrac{30000-xy}{3 ...<math>\sqrt{3}\tan{\left(\alpha\right)}</math>, we can set <math>\sqrt{3}\tan{\left(\alpha\right)}=a</math> for convenience since we really only care abo
  15 KB (2,560 words) - 01:44, 1 July 2023
• 1983 AHSME Problems/Problem 20
  If <math>\tan{\alpha}</math> and <math>\tan{\beta}</math> are the roots of <math>x^2 - px + q = 0</math>, and <math>\co ...cot\theta=\frac{1}{\tan\theta}</math>, we have <math>\frac{1}{\tan(\alpha)\tan(\beta)}=\frac{1}{q}=s</math>.
  1 KB (222 words) - 00:58, 20 February 2019
• 2016 USAMO Problems/Problem 3
  ...= \frac {VI_A}{VO} = \frac {R \sin \psi + 2R \sin \alpha}{R \cos \psi} = \tan \psi + \frac{2 \sin\alpha}{\cos \psi}.</cmath> ...cot \angle UTW = \frac {TW}{UW} = \frac {AW \cdot \tan \psi}{AU – AW} = \tan \psi \cdot \frac {2a +b+c}{b+c} =</cmath>
  6 KB (998 words) - 21:36, 17 October 2022
• 2016 USAMO Problems/Problem 5
  ...s clear that <math>I = \left(\frac{b + c – a}{2} , \frac{b + c – a}{2}\tan(A / 2)\right)</math>. ...and the <math>y</math> coordinate of <math>O</math> is <math>-\frac{b}{2} \tan{B-90}</math>. From this, <math>(5)</math> follows in this case as well.
  8 KB (1,449 words) - 00:09, 12 October 2023
• 1983 AHSME Problems/Problem 5
  ...angle at <math>C</math>. If <math>\sin A = \frac{2}{3}</math>, then <math>\tan B</math> is so <math>\tan{B} = \frac{x \sqrt{5}}{2x} = \frac{\sqrt{5}}{2}</math>, which is choice <ma
  1 KB (171 words) - 00:42, 20 February 2019
• 1979 AHSME Problems/Problem 20
  ...{2}+\frac{1}{3}}{1-(\frac{1}{2})(\frac{1}{3})}</math>. Simplifying, <math>\tan(\theta_a + \theta_b) = 1</math>, so <math>\theta_a + \theta_b</math> in rad
  2 KB (363 words) - 12:46, 10 May 2022
• 2017 AMC 12A Problems/Problem 15
  Let <math>f(x) = \sin{x} + 2\cos{x} + 3\tan{x}</math>, using radian measure for the variable <math>x</math>. In what in ...tan function. Upon further examination, it is clear that the positive the tan function creates will balance the other two functions, and thus the first s
  3 KB (564 words) - 14:12, 23 October 2021
• 2017 AMC 12A Problems
  Let <math>f(x) = \sin{x} + 2\cos{x} + 3\tan{x}</math>, using radian measure for the variable <math>x</math>. In what in
  15 KB (2,418 words) - 16:58, 7 November 2022
• 2017 AIME I Problems/Problem 15
  <cmath>x(\frac {\sin \beta}{\tan{\alpha}} - \cos \beta) +x (\frac {\cos\beta}{2} +\frac{\sin\beta \sqrt{3}}{
  22 KB (3,622 words) - 17:11, 6 January 2024
• Reaper Archives
  | 67 || Senpai-Tan || 80 || 8151.479 || 101.893
  187 KB (10,824 words) - 18:27, 3 February 2022
• 1972 AHSME Problems/Problem 20
  If <math>\tan x=\dfrac{2ab}{a^2-b^2}</math> where <math>a>b>0</math> and <math>0^\circ <x We start by letting <math>\tan x = \frac{\sin x}{\cos x}</math> so that our equation is now: <cmath>\frac{
  1 KB (177 words) - 19:14, 2 January 2024
• 1972 AHSME Problems/Problem 30
  real x = 6-h*tan(t); real y = x*tan(2*t);
  3 KB (431 words) - 19:52, 23 June 2021
• 2020 AMC 12A Problems
  How many solutions does the equation <math>\tan{(2x)} = \cos{(\tfrac{x}{2})}</math> have on the interval <math>[0, 2\pi]?</
  14 KB (2,073 words) - 15:15, 21 October 2021
• 2017 AMC 8 Problems/Problem 22
  ...an use the famous mnemonic SOH CAH TOA. <math>AD=AB-DB=13-5=8 \Rightarrow \tan \angle BAC = \frac{5}{12}=\frac{r}{8} \Rightarrow 12r=40 \Rightarrow r= \fr
  5 KB (762 words) - 21:20, 20 January 2024
• MIE 96/97
  If <math>\tan a</math> and <math>\tan b</math> are the roots of <math>x^2+px+q=0</math>, then compute, in terms o
  2 KB (377 words) - 14:52, 7 January 2018
• MIE 2016/Day 1/Problem 4
  &\tan(2b)= &\frac{1}{4}\\&
  571 bytes (90 words) - 05:19, 17 June 2021
• MIE 2015
  <math>\sin x\left(1+\tan x\tan\frac{x}{2}\right)=4-\cot x</math>
  7 KB (1,127 words) - 18:23, 11 January 2018
• British Mathematical Olympiad
  ...> height of the cone be <math>h,</math> radius of the cone be <math>r = h \tan \theta.</math> <cmath>BO = a, BC = \frac {a}{\sqrt {2}}, AO = h, DO = r = h \tan \theta.</cmath>
  6 KB (1,034 words) - 10:12, 7 June 2023
• 2018 AMC 12A Problems/Problem 23
  ...\frac{2\sin(46)\cos(10)}{-2\sin(46)\sin({-10})}=\frac{\sin(80)}{\cos(80)}=\tan(80)</cmath>
  12 KB (1,878 words) - 22:11, 23 October 2021
• 2018 AMC 12A Problems/Problem 22
  ...c{7\pi}{6}</math> without the loss of generality. Since <math>\tan(2\phi)>\tan\frac{\pi}{3},</math> we deduce that <math>2\phi>\frac{\pi}{3},</math> from
  10 KB (1,662 words) - 12:45, 13 September 2021
• 1969 IMO Problems/Problem 2
  <cmath>\tan{x_1}=\frac{\cos{a_1}+\frac{1}{2}\cos{a_2}+\frac{1}{4}\cos{a_3}+\cdots+\frac .../math> is <math>\pi</math>, this means that <math>\tan{x_1}=\tan{x_1+\pi}=\tan{x_1+m\pi}</math> for any natural number <math>m</math>. That implies that e
  1 KB (269 words) - 11:29, 4 April 2024
• 2018 AIME I Problems/Problem 13
  <cmath>\frac {(r-r_1)\cdot (r-r_2)}{r_1 \cdot r_2} =\tan\beta \tan\gamma.</cmath> <cmath>1 -\frac{2r}{h} = \frac {b+c-a}{b+c+a} = \frac {r}{r_a} = \tan\beta \tan\gamma .</cmath>
  13 KB (2,200 words) - 21:36, 6 January 2024
• 1973 AHSME Problems/Problem 17
  ...nd <math> \sin \frac12 \theta = \sqrt{\frac{x-1}{2x}}</math>, then <math> \tan \theta</math> equals <cmath>\tan \frac{\theta}{2} = \sqrt{\frac{x-1}{2x}} \div \sqrt{\frac{x+1}{2x}}</cmath>
  1 KB (184 words) - 14:00, 20 February 2020
• 1983 AHSME Problems/Problem 27
  ...}{2}\right )\right )=\tan \left (\frac{1}{2} \right )</math>. Since <math>\tan \left(\frac{\theta}{2} \right ) = \frac{1-\cos \left(\theta \right )}{\sin
  1 KB (245 words) - 14:00, 29 January 2023
• 2008 iTest Problems/Problem 46
  ...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>. ...0</math>. By the Zero Product Property, <math>\tan x = 0</math> or <math>\tan x = 2\sin x</math>.
  969 bytes (158 words) - 19:00, 12 July 2018
• 2008 iTest Problems/Problem 57
  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 ...the sum formula for tangent, the sum of the two possible values of <math>\tan \theta</math> is
  2 KB (343 words) - 20:35, 4 August 2018
• 2006 Indonesia MO Problems
  ...all triangles <math> ABC</math> which have property: <math> \tan A,\tan B,\tan C</math> are positive integers. Prove that all triangles in <math> S</math>
  3 KB (439 words) - 12:39, 4 September 2018
• 2008 iTest Problems/Problem 96
  \tan(2x) &= \frac{\sqrt{3}}{4}.
  3 KB (558 words) - 20:13, 4 January 2019
• 2008 iTest Problems/Problem 98
  \tan(\angle AIS + \angle DIS) &= -\tan(\angle BIT + \angle CIT) \\ ...ngle DIS} &= \frac{\tan \angle BIT + \tan \angle CIT}{1 - \tan \angle BIT \tan \angle CIT}
  3 KB (586 words) - 23:47, 8 January 2019
• 2019 AIME II Problems/Problem 10
  ...th> such that for nonnegative integers <math>n</math>, the value of <math>\tan{\left(2^{n}\theta\right)}</math> is positive when <math>n</math> is a multi Note that if <math>\tan \theta</math> is positive, then <math>\theta</math> is in the first or thir
  7 KB (1,109 words) - 00:40, 28 January 2024
• 2019 USAJMO Problems/Problem 4
  <cmath>FG = AG - FG = \frac{r_a}{\tan \left( \frac{A}{2} \right)} - b \cos (A)</cmath> <cmath>FH = AH - AF = \frac{r_a}{\tan \left( \frac{A}{2} \right)} - c \cos (A)</cmath>
  10 KB (1,536 words) - 20:27, 12 April 2021
• 2006 iTest Problems/Problem U6
  ...ouble Angle Identity yields <math>\tan 2\theta = \frac34</math>, so <math>\tan (90 - 2\theta) = \frac43</math>.
  4 KB (722 words) - 20:53, 27 March 2019
• 2019 AMC 10B Problems/Problem 23
  ...olving, we obtain <math>\tan{x}=\frac{1}{4}</math>. Then, note that <math>\tan{x}=r/{BC}</math>, so <math>r=\frac{1}{4}*\sqrt{170}</math>. Finishing off,
  6 KB (967 words) - 10:25, 20 December 2023
• 2019 AIME I Problems/Problem 6
  ...65}{28}=\frac{28\sin\alpha}{8\cos\alpha}=\frac{7}{2}\tan\alpha\Rightarrow \tan\alpha=\frac{65}{98}</cmath> Thus, <math>MK=\frac{MN}{\tan\alpha}=98</math>, so <math>MO=MK-KO=\boxed{090}</math>.
  11 KB (1,717 words) - 20:11, 19 January 2024
• 2019 AIME II Problems
  ...th> such that for nonnegative integers <math>n,</math> the value of <math>\tan(2^n\theta)</math> is positive when <math>n</math> is a multiple of <math>3<
  7 KB (1,254 words) - 14:45, 21 August 2023
• 2019 AIME II Problems/Problem 1
  <cmath>\tan a = \frac{8}{15}</cmath> <cmath>A = \frac{1}{2}* 9*\frac{9}{2}\tan a = \frac{54}{5}</cmath>
  11 KB (1,794 words) - 15:32, 14 January 2024
• Euler's formula and It's history
  x = tan-1 ( 4 / 3 ) = 0.927 (to 3 decimals)
  3 KB (543 words) - 15:24, 13 June 2019
• Mock AIME 2 Pre 2005 Problems
  ...lpha) + \tan^{-1}(\beta) + \tan^{-1} (\gamma).</cmath> The value of <math>\tan(\omega)</math> can be written as <math>\tfrac{m}{n}</math> where <math>m</m
  6 KB (1,052 words) - 13:52, 9 June 2020
• 2006 SMT/Calculus Problems/Problem 5
  Evaluate: <math> \int(x\tan^{-1}x)dx </math> \int(x\tan^{-1}x)dx &= \frac{x^2}{2}\tan^{-1}x-\int\frac{x^2}{2(x^2+1)}dx\\
  670 bytes (116 words) - 18:31, 14 January 2020
• 2006 SMT/Advanced Topics Problems/Problem 10
  Using the identity that <math>\tan(x) = -\tan(-x)</math>
  1 KB (175 words) - 18:30, 14 January 2020
• 2006 Indonesia MO Problems/Problem 3
  ...all triangles <math> ABC</math> which have property: <math> \tan A,\tan B,\tan C</math> are positive integers. Prove that all triangles in <math> S</math> ...B+C = 180^\circ</math>, so <math>z = \tan C = \tan (180^\circ - (A+B)) = -\tan(A+B)</math>.
  3 KB (465 words) - 12:00, 26 September 2019
• 2020 AIME I Problems/Problem 13
  We compute that <math>\cos{\angle{ABC}}=\frac{1}{8}</math>, so <math>\tan{\angle{ABC}}=3\sqrt{7}</math>. ...n \angle ABC}{1 + \cos \angle ABC} = \frac{\sqrt{7}}{3}</math>, and <math>\tan \angle DAF = \frac{\sqrt{7}}{7}</math>.
  35 KB (5,215 words) - 23:08, 29 October 2023
• 2023 AIME I Problems/Problem 12
  ...erty that <cmath>\angle AEP = \angle BFP = \angle CDP.</cmath> Find <math>\tan^2(\angle AEP).</math> ...ath>D=(0, 0)</math>, and <math>C=(48, 0)</math>, where we will find <math>\tan^{2}\left(\measuredangle CDP\right)</math> with <math>P=(BFD)\cap(CDE)</math
  16 KB (2,592 words) - 15:40, 13 April 2024
• 2000 Pan African MO Problems/Problem 1
  <cmath> \sin^3{x}(1+\cot{x})+\cos^3{x}(1+\tan{x})=\cos{2x} </cmath> ...= 0</math>, we can divide both sides by <math>\cos{x}</math> to get <math>\tan{x} = -1</math>. Thus, <math>x = \frac{3 \pi}{4} + \pi n</math>, where <mat
  2 KB (305 words) - 06:07, 23 February 2023
• 2006 iTest Problems/Problem U7
  ...opposite angle to <math>x</math> be <math>\theta</math>, and let <math>t:=\tan\frac{\theta}{2}</math>; let the [[area]] be <math>A</math> and the [[semipe
  4 KB (674 words) - 16:03, 25 February 2021
• 2000 Pan African MO Problems
  <cmath> \sin^3{x}(1+\cot{x})+\cos^3{x}(1+\tan{x})=\cos{2x} </cmath>
  2 KB (393 words) - 13:39, 4 December 2019
• 2000 SMT/Calculus Problems
  ...t value of <math>x</math> <math>(0 < x < \frac{\pi}{2})</math> does <math>\tan x + \cot x</math> achieve its minimum?
  3 KB (413 words) - 13:10, 21 January 2020
• 2000 SMT/Team Problems
  ...-\beta) = 0</math>, <math>\tan(\beta) = \frac{1}{2000}</math>, find <math>\tan(\alpha)</math>.
  4 KB (618 words) - 13:33, 21 January 2020
• Unit Circle
  ...t can be used to demonstrate trigonometric functions such as sin, cos, and tan, but it is most commonly used to visualize the complex numbers. This is don
  741 bytes (131 words) - 11:50, 22 January 2020
• 2020 AMC 10A Problems/Problem 20
  ...lve the quadratic, taking the positive solution (C is acute) to get <math>\tan{C} = \frac{1}{3}.</math> So if <math>AB = a,</math> then <math>BC = 3a</mat ...n angle bisector of <math>\triangle ABC</math> (because we will get <math>\tan(x) = 1</math>).
  13 KB (2,046 words) - 18:33, 28 October 2023
• 2020 AMC 12A Problems/Problem 9
  How many solutions does the equation <math>\tan(2x)=\cos(\tfrac{x}{2})</math> have on the interval <math>[0,2\pi]?</math> We count the intersections of the graphs of <math>y=\tan(2x)</math> and <math>y=\cos\left(\frac x2\right):</math>
  4 KB (615 words) - 04:07, 8 July 2022
• 2020 AMC 12A Problems/Problem 12
  ...\frac{3}{5}</math>, and the angle we are rotating around is x, then <math>\tan x = \frac{3}{5}</math> <math>\tan(x+45^{\circ}) = \frac{\tan x + \tan(45^{\circ})}{1-\tan x*\tan(45^{\circ})} = \frac{0.6+1}{1-0.6} = \frac{1.6}{0.4} = 4</math>
  7 KB (1,145 words) - 20:27, 5 November 2023
• 2020 AMC 12A Problems/Problem 14
  ...}{2}</math>, the area of the octagon is then <math>\frac{1}{2} \cdot \text{tan}(67.5) \cdot \frac{8}{2}</math>. = 4 \cdot \frac{\text{sin}^2(67.5)}{2\cdot \text{tan}(67.5)}
  11 KB (1,654 words) - 02:01, 17 September 2023
• 2020 AMC 12B Problems/Problem 7
  ==Solution 4 (tan)== ...45^{\circ}+\theta)=\frac{\tan(45^{\circ})+\tan(\theta)}{1-\tan(45^{\circ})\tan(\theta)} = \frac{1+a}{1-a}</math>. Since the slope of one line is <math>6</
  5 KB (895 words) - 15:03, 8 June 2023
• 2020 AIME I Problems/Problem 1
  A = (0, tan(3 * pi / 7));
  6 KB (968 words) - 15:01, 24 January 2024
• AMC 12C 2020
  How many solutions does the trigonometric equation <math>tan(cos(x)) = cos((x\pi^{2} - sin(x))</math> have in the interval <math>[-\pi,
  9 KB (1,450 words) - 18:33, 21 April 2020
• Pythagorean identities
  <math>\tan^2x + 1 = \sec^2x</math>
  975 bytes (164 words) - 14:07, 3 January 2024
• Angle addition identities
  <math>\tan(x + y) = \frac{\tan (x) + \tan (y)}{1 - \tan (x) \tan (y)}</math> ...cos \alpha \cos \beta}} = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}</math>
  2 KB (372 words) - 20:46, 13 January 2024
• Double angle identities
  * <math>\tan (2x) = \frac{2\tan (x)}{1-\tan^2 (x)}</math>
  357 bytes (54 words) - 20:21, 7 September 2023
• Half angle identities
  * <math>\tan \left(\frac{x}{2}\right) = \pm \sqrt{\frac{1 - \cos (x)}{1+\cos (x)}} = \fr
  956 bytes (149 words) - 20:23, 7 September 2023
• 2020 AIME II Problems/Problem 4
  ...n</math> angle formula, <math>\tan{(a-b)}=\frac{\tan(a)-\tan(b)}{1+\tan(a)\tan(b)}</math> to find the slope of line <math>CP</math>. We know that line <ma
  10 KB (1,542 words) - 13:29, 19 January 2024
• 2020 AIME II Problems/Problem 13
  ...<math>\tan \left(\frac{\angle B}{2}\right) = \frac{r}{2}</math> and <math>\tan \left(\frac{\angle E}{2}\right) = \frac{r}{4}</math>, so <cmath>\cos (\angle B) =\frac{1-\tan^2 \left(\frac{\angle B}{2}\right)}{1+\tan^2 \left(\frac{\angle B}{2}\right)} = \frac{4-r^2}{4+r^2}</cmath>and
  13 KB (2,197 words) - 23:00, 8 January 2024
• 2021 AIME I Problems/Problem 2
  ...ngles from the larger rectangle, we get  Area = <math>33-3BG=33-\frac{9}{\tan(\angle DAE)}</math>. <math>\alpha=\tan^{-1}\left(\frac{3}{11}\right)</math>
  8 KB (1,294 words) - 00:59, 23 August 2022
• 2021 AIME I Problems/Problem 11
  The angle <math>\theta</math> between diagonals satisfies <cmath>\tan{\frac{\theta}{2}}=\sqrt{\frac{(s-b)(s-d)}{(s-a)(s-c)}}</cmath> (see https:/ ...\tan{\frac{\theta}{2}}=\sqrt{\frac{(11-4)(11-6)}{(11-5)(11-7)}}\text{ or }\tan{\frac{\theta}{2}}=\sqrt{\frac{(11-5)(11-7)}{(11-4)(11-6)}}.</cmath>
  16 KB (2,635 words) - 19:56, 24 December 2023
• Mock AIME 2 Pre 2005 Problems/Problem 11
  ...lpha) + \tan^{-1}(\beta) + \tan^{-1} (\gamma).</cmath> The value of <math>\tan(\omega)</math> can be written as <math>\tfrac{m}{n}</math> where <math>m</m ...ma - \gamma\alpha}</math>. Substituting Vieta's formulas, we obtain <math>\tan(\omega) = \frac{\frac{799}{4} - \frac{1}{4}}{1 - (-50)} = \frac{\frac{798}{
  1 KB (192 words) - 18:03, 13 September 2020
• 2023 AIME II Problems
  .../math> less than or equal to <math>1000</math> such that <math>\sec^n A + \tan^n A</math> is a positive integer whose units digit is <math>9.</math>
  8 KB (1,370 words) - 21:34, 28 January 2024
• 2022 AMC 12B Problems
  .../math> and the <math>y</math>-axis at point <math>B</math>. What is <math>\tan(\angle ABC)</math>?
  15 KB (2,233 words) - 13:02, 10 November 2023
• 2021 Fall AMC 10B Problems/Problem 25
  Hence, <math>\tan \theta = \frac{1}{3}</math>.
  16 KB (2,274 words) - 09:02, 10 December 2022
• 2021 AIME II Problems
  ...the acute angle formed by the diagonals of the quadrilateral. Then <math>\tan \theta</math> can be written in the form <math>\tfrac{m}{n}</math>, where <
  8 KB (1,429 words) - 14:31, 26 February 2024
• 2021 AMC 10A Problems/Problem 24
  Let <math>\tan A=a.</math> The area of the rectangle created by the four equations can be
  10 KB (1,662 words) - 19:31, 18 November 2022
• 1978 AHSME Problems/Problem 15
  If <math>\sin x+\cos x=1/5</math> and <math>0\le x<\pi</math>, then <math>\tan x</math> is Since <math>\tan x = \frac{\sin x}{\cos x}</math>, we have
  1 KB (207 words) - 21:10, 13 February 2021
• 2021 AIME II Problems/Problem 12
  ...the acute angle formed by the diagonals of the quadrilateral. Then <math>\tan \theta</math> can be written in the form <math>\tfrac{m}{n}</math>, where < ...and <math>\cos \theta</math> separately and use their values to get <math>\tan \theta</math>. We can start by drawing a diagram. Let the vertices of the q
  10 KB (1,669 words) - 17:33, 12 January 2024
• 2022 AIME II Problems/Problem 15
  <cmath>\tan \alpha =\frac { \sin 2 \alpha}{1+\cos 2 \alpha} = \frac {4/5}{1 - 3/5}=2.</ The radius of the inscribed circle is <math>r = (s – B'D) \tan \alpha.</math>
  14 KB (2,217 words) - 00:28, 29 June 2023
• 2021 Fall AMC 12B Problems/Problem 15
  ...^\circ</math>, so <math>\angle{EAB} = 30^{\circ}</math>. Then <math>EB=AE\tan 30^\circ = \sqrt{3}</math>; therefore <math>BC=EC-EB=3-\sqrt{3}</math>. Thu
  4 KB (686 words) - 18:38, 12 July 2023
• 2021 Fall AMC 12B Problems/Problem 22
  ...d <math>MP\perp AB</math>. Let <math>\theta = \angle BAC</math>; so <math>\tan\theta = \tfrac 68 = \tfrac 34</math>. Then <cmath>OP=MP-MO=AM\cot\theta - BM\tan\theta = 5(\tfrac 43 - \tfrac 34) = \boxed{\textbf{(C)}\ \tfrac{35}{12}}.</c
  6 KB (943 words) - 00:41, 6 August 2023
• 2021 Fall AMC 12B Problems/Problem 9
  ...erp</math> <math>\overline{AB}</math> and <math>AY = 3</math>. Thus <math>\tan(30^\circ) = \frac{OY}{3}</math> so <math>OY = \sqrt{3}</math>.
  5 KB (765 words) - 22:33, 10 July 2023
• 2021 Fall AMC 10A Problems/Problem 19
  ~MathFun1000 (Inspired by Way Tan)
  3 KB (475 words) - 17:45, 30 October 2023
• 2021 Fall AMC 10B Problems/Problem 11
  ...could find the length of the apothem by the formula <math>\frac{s}{2\text{tan}(\frac{180}{n})},</math> where <math>s</math> is the side length and <math>
  6 KB (1,021 words) - 19:40, 1 November 2023
• 2021 Fall AMC 10B Problems/Problem 17
  ...th the positive x axis. Thus, line <math>m</math> makes an angle of <math>\tan^{-1}(5) - 45^\circ</math> with the positive x axis. Thus, the slope of line <cmath> \tan (\tan^{-1}(5) - 45^\circ) = \frac{5 - 1}{1 + 5\cdot 1} = \frac{2}{3},</cmath>
  8 KB (1,331 words) - 22:44, 16 December 2023
• 2021 Fall AMC 12A Problems/Problem 13
  ...th>\theta_{3}</math>. Note that <math>\tan(\theta_{1})=1</math> and <math>\tan(\theta_{3})=3</math>, and this is why we named them as such. Let the angle \tan(\theta_k-\theta_1)&=\tan(\theta_3-\theta_k)\\
  16 KB (2,526 words) - 00:53, 6 May 2023
• 2023 AIME I Problems/Problem 5
  ...rom <math>P</math> to line <math>\overline{BD}</math>. We know that <math>\tan{\angle{POX}} = \frac{PX}{XO} = \frac{28}{45}</math> by triangle area and gi \frac{(2 \cdot \tan{\angle{OCP}})}{(1-\tan^2{\angle{OCP}})} = \tan{\angle{POX}} = \frac{28}{45}
  19 KB (3,107 words) - 23:31, 17 January 2024
• 2023 AIME I Problems
  ...erty that <cmath>\angle AEP = \angle BFP = \angle CDP.</cmath> Find <math>\tan^2(\angle AEP).</math>
  7 KB (1,154 words) - 12:54, 20 February 2024
• Kimberling’s point X(24)
  <cmath>HB' = 2 HD \sin (\alpha - 90^\circ) = - 2 CD \tan(90^\circ- \beta) \cos \alpha = - 2 AC \cos \gamma \frac {\cos \beta}{\sin \ <cmath>LM = \frac {R}{2} (\tan \beta + \tan \gamma) = \frac {R \sin (\beta + \gamma)}{2 \cos \beta \cdot \cos \gamma} \
  10 KB (1,751 words) - 15:34, 25 November 2022
• 2022 AMC 10B Problems/Problem 20
  By solving this equation, we get <math>\tan \angle FBC = \frac{\sin 46^\circ}{2 + \cos 46^\circ}</math>. \tan \angle BFC & = \frac{\sin \angle FBC}{\cos \left( 46^\circ - \angle FBC \ri
  11 KB (1,742 words) - 20:06, 8 September 2023
• 2022 AMC 12B Problems/Problem 25
  <cmath>\frac 14\mathrm{base length}^2\cdot\tan(\mathrm{base angle})=\frac 14((\sqrt3-1)\sqrt2)^2\cdot\tan15^\circ=\frac 14
  20 KB (2,980 words) - 18:17, 2 January 2024
• 2022 AMC 12B Problems/Problem 14
  .../math> and the <math>y</math>-axis at point <math>B</math>. What is <math>\tan(\angle ABC)</math>? Thus, <cmath>\tan(\angle ABC) = \sqrt{\frac{65}{49}-1}= \boxed{\textbf{(E)}\ \frac{4}{7}}.</c
  7 KB (1,013 words) - 22:23, 27 October 2023
• Gossard perspector
  ...\tan \theta_C = \frac{3 – \tan \beta \cdot \tan \alpha}{\tan \beta – \tan \alpha}.</math> ...e O'KF = \frac{3 – \tan \theta_B \cdot \tan \theta_C}{\tan \theta_C – \tan \theta_B}.</math>
  18 KB (3,046 words) - 06:44, 19 January 2023
• 2023 AIME I Problems/Problem 8
  Which also gives us <math>\tan{\angle{OWX}}=\frac{1}{2}</math> and <math>OW=\frac{25\sqrt{5}}{2}</math>. \frac{OX}{OW}&=\tan{\angle{OWX}} \\
  17 KB (2,612 words) - 14:54, 3 July 2023
• 2023 AIME I Problems/Problem 13
  Denote <math>\alpha = \tan^{-1} \frac{\sqrt{21}}{\sqrt{31}}</math>. & = \pm d \cos 2 \alpha \tan \alpha .
  13 KB (2,130 words) - 01:52, 31 January 2024
• 2023 AIME II Problems/Problem 13
  .../math> less than or equal to <math>1000</math> such that <math>\sec^n A + \tan^n A</math> is a positive integer whose units digit is <math>9.</math> Denote <math>a_n = \sec^n A + \tan^n A</math>.
  5 KB (881 words) - 04:52, 19 December 2023
• 2023 AIME II Problems/Problem 12
  <cmath>\tan(2\angle{MAL_1}+\angle{L_1AB}-\angle{CAL_1})=\tan(\angle{QBL_2}-\angle{QCL_2})</cmath> ...}{6}</math>, <math>\tan{CAL_1}=\frac{9}{12}=\frac{3}{4}</math>, and <math>\tan{L_1AB}=\frac{5}{12}</math>. By abusing the tangent angle addition formula,
  12 KB (1,900 words) - 18:14, 28 January 2024
• 2023 AIME II Problems/Problem 14
  <cmath>\tan \alpha = \frac {\sin \alpha}{\cos \alpha} = \frac {h(B)}{AB}\cdot \frac {AC
  10 KB (1,574 words) - 16:42, 8 March 2024
• 2023 AMC 12A Problems
  ...a_{2023} \tan^{2023} x}{1 + a_2 \tan^2 x + a_4 \tan^4 x \cdots + a_{2022} \tan^{2022} x} </cmath>whenever <math>\tan 2023x</math> is defined. What is <math>a_{2023}?</math>
  15 KB (2,168 words) - 05:11, 4 February 2024
• 2023 AMC 12A Problems/Problem 11
  .../math>. <math>\tan(\alpha-\beta)=\dfrac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}=\dfrac{2-1/3}{1+2\cdot 1/3}=1</math>. Therefore, <math>\alpha-\beta=\
  6 KB (907 words) - 23:48, 18 February 2024
• 2023 AMC 12A Problems/Problem 25
  ...a_{2023} \tan^{2023} x}{1 + a_2 \tan^2 x + a_4 \tan^4 x \cdots + a_{2022} \tan^{2022} x} </cmath>whenever <math>\tan 2023x</math> is defined. What is <math>a_{2023}?</math>
  6 KB (960 words) - 01:27, 11 November 2023
• 2023 AMC 12A Problems/Problem 9
  ...theta</math> be the angle opposite the smaller leg. We want to find <math>\tan\theta</math>. ...ta = \frac{1}{2},</math> or <math>\theta=15^\circ</math>. Therefore <math>\tan \theta = \boxed{\textbf{(C) }2-\sqrt3}</math>
  7 KB (1,106 words) - 09:10, 8 March 2024
• 2023 AMC 12A Problems/Problem 15
  &= \frac{100 - \frac{90}{\tan\theta}}{120 - \frac{90}{\sin\theta}}\\
  7 KB (1,074 words) - 21:22, 20 November 2023
• 2023 AMC 12B Problems/Problem 25
  ...\frac{s}{2} = \frac{s}{4} \cdot (sec^2(54^{\circ})-2) = \frac{s}{4} \cdot (tan^2(54^{\circ})-1)</math>. ...t in a right triangle. Thus, <math>FU = \frac{s}{4} \cdot (tan(54^{\circ})-tan(36^{\circ}))</math>
  19 KB (2,967 words) - 16:56, 24 February 2024
• Polya’s method for extremums
  ...t \sin \alpha = F_B \cdot BC \cdot \cos \alpha \implies \frac {BC}{AC} = \tan^2 \alpha,</cmath> ...h>\frac {y_C}{x_C} = \frac {BC \cdot \sin \alpha}{AC \cdot \cos \alpha} = \tan^3 \alpha \implies</cmath>
  12 KB (2,104 words) - 14:11, 24 February 2024
• 2024 AIME I Problems/Problem 7
  ...the fourth quadrant (side lengths of <math>3, -4, 5</math>). Since <math>\tan\theta = -\frac{4}{3},</math> we quickly see that <math>\sin\theta = -\dfrac
  12 KB (1,842 words) - 19:26, 23 February 2024
• Bisector
  ...{MG}{MF} = \frac {AM \tan \alpha}{BM \tan \gamma} =\frac {\tan \alpha}{\tan \gamma} = \frac {CB''}{AC''}=\frac{a+b-c}{b+c-a}.</cmath> ...^2 \alpha} = \frac {\sin 2\gamma}{\sin 2\alpha} \cdot \frac {\tan \gamma}{\tan \alpha} = \frac {c}{a} \cdot \frac{b+c-a}{a+b-c}.</cmath>
  10 KB (1,668 words) - 09:47, 26 March 2024
• 1991 OIM Problems/Problem 2
  <math>A_1=x\frac{y-x.tan(\theta)+y}{2}+\frac{y^2tan(\theta)}{2}</math> <math>A_1=\frac{y^2-x^2}{2}tan(\theta)+xy=1</math>
  3 KB (482 words) - 09:40, 23 December 2023
• 2023 SSMO Relay Round 1 Problems
  <cmath>\sec^{N} (Nx) - \tan^{N}(Nx) = 1</cmath>
  640 bytes (94 words) - 21:32, 15 December 2023
• 2023 SSMO Relay Round 1 Problems/Problem 3
  <cmath>\sec^{N} (Nx) - \tan^{N}(Nx) = 1</cmath>
  180 bytes (31 words) - 22:28, 15 December 2023
• 2024 AIME I Problems/Problem 5
  Let <math>\angle{DHE} = \theta.</math> This means that <math>DE = 17\tan{\theta}.</math> Since quadrilateral <math>ADHG</math> is cyclic, <math>\ang ...a} = \frac{16}{DX} = \frac{17}{FX}</math> and <math>DX + FX = DE + EF = 17\tan{\theta} + 184.</math>
  7 KB (1,144 words) - 19:25, 23 February 2024
• Proofs of trig identities
  <math>\tan = \frac{\sin}{\cos}</math> pair D = (1,tan(d));
  16 KB (2,796 words) - 13:12, 21 January 2024
• 2024 AIME II Problems/Problem 15
  Case 1: <math>k = 0, \tan 30^\circ, \tan 60^\circ</math>. ...th>. The number of solutions for <math>k = \tan 30^\circ</math> and <math>\tan 60^\circ</math> are the same.
  8 KB (1,395 words) - 17:26, 9 February 2024
• DVI exam
  <cmath>\tan \angle MSN = \frac {1}{2\sqrt{3}} \implies</cmath> <cmath>AM = KM \cdot \tan \alpha, BM = \frac {KM}{\tan \alpha},</cmath>
  19 KB (3,295 words) - 04:11, 12 April 2024
• 2024 AIME II Problems/Problem 12
  ...=-\sqrt{3}x+\frac{\sqrt{3}}{2}, x=\frac{\sqrt{3}-2\sin \theta}{2\sqrt{3}-2\tan \theta}</math> ...math>\lim_{\theta\to\frac{\pi}{3}}\frac{\sqrt{3}-2\sin \theta}{2\sqrt{3}-2\tan \theta}=\lim_{\theta\to\frac{\pi}{3}}cos^3{x}=\frac{1}{8}</math>. This mean
  10 KB (1,655 words) - 00:31, 11 April 2024
• Sharygin Olympiads, the best
  <cmath>\angle CBO = 90^\circ \implies \angle COB = \alpha, MB = b \tan \alpha,</cmath> Denote <math>Q = AB \cap DP \implies BQ = \frac {BP}{\cos \alpha} = b \tan \alpha = MB.</math>
  19 KB (3,323 words) - 06:20, 15 April 2024