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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

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