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• 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) - 13:33, 16 July 2024
• 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
• 2013 Mock AIME I Problems/Problem 12
  ...7{AE/2} = \tfrac{14}{AE}</math>. Equating these two expressions for <math>\tan\theta</math>, we see that <math>AE \cdot CE = 7 \cdot 14 = 98</math>. From
  6 KB (1,035 words) - 10:42, 31 July 2024
• 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
• 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) - 21:28, 6 August 2024
• 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 ...ath> 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
  5 KB (1,005 words) - 00:55, 9 August 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

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