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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
2009 AMC 10A Problems/Problem 10
...get <math>xy = 3</math>, and using the fact that <math>\cot{a} = \frac{1}{\tan{a}}</math>, <math>\frac{x}{y} = 4</math>. Solving for x and y(using substit

5 KB (879 words) - 18:57, 30 April 2024
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 (318 words) - 13:47, 4 May 2024
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

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