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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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