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  • 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
  • ...>. 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
  • | <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
  • *<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
  • & = &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
  • ...}{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
  • ...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
  • ...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
  • ...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
  • \begin{matrix} {CE} & = & r \tan(COE) \\
    4 KB (684 words) - 07:28, 3 October 2021
  • ...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
  • 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
  • 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
  • \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
  • \tan{\alpha}=\frac{4nh}{(n^2-1)a}.
    3 KB (511 words) - 21:21, 20 August 2020
  • ...{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
  • ...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
  • ...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
  • <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
  • ...||\cos||<math>\textstyle \sin</math>||\sin||<math>\textstyle \tan</math>||\tan
    12 KB (1,898 words) - 15:31, 22 February 2024

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