Create the page "Tan" on this wiki! See also the search results found.

• 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

View (previous 20 | next 20) (20 | 50 | 100 | 250 | 500)