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• 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) - 17:43, 22 April 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 \alpha = \frac {SO}{OH} = \frac {GO}{QH} \implies OH \cdot GO = \frac {15} <cmath>h = 5 \sqrt {\frac {3}{2}} \implies \tan \alpha = \sqrt{2} \implies \cos \alpha = \frac {1}{\sqrt{3}}.</cmath>
  32 KB (5,375 words) - 11:43, 5 May 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>
  23 KB (4,003 words) - 16:17, 21 April 2024
• 2005 iTest Problems
  ...ath> equal the correct answer from 3A. In triangle <math>WXY</math>, <math>tan \angle YWX= (A + 8) / .5A</math>, and the altitude from <math>W</math> divi
  25 KB (3,738 words) - 10:53, 23 April 2024

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