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  • ...we make use of the identity <math>\tan^2x+1=\sec^2x</math>. Set <math>x=a\tan\theta</math> and the radical will go away. However, the <math>dx</math> wil Since <math>\sec^2(\theta)-1=\tan^2(\theta)</math>, let <math>x=a\sec\theta</math>.
    1 KB (173 words) - 18:42, 30 May 2021
  • * <math>\tan^2x + 1 = \sec^2x</math> * <math>\tan(x + y) = \frac{\tan (x) + \tan (y)}{1 - \tan (x) \tan (y)} </math>
    8 KB (1,371 words) - 10:33, 7 November 2021
  • ...de opposite <math>A</math> to the side adjacent to <math>A</math>. <cmath>\tan (A) = \frac{\textrm{opposite}}{\textrm{adjacent}} = \frac{a}{b}.</cmath> ...e reciprocal of the tangent of <math>A</math>. <cmath>\cot (A) = \frac{1}{\tan (x)} = \frac{\textrm{adjacent}}{\textrm{opposite}} = \frac{b}{a}.</cmath>
    8 KB (1,217 words) - 11:44, 4 March 2022
  • | Tan
    59 KB (7,185 words) - 17:56, 2 July 2022
  • <math>\textbf {(A)}\ \sec^2 \theta - \tan \theta \qquad \textbf {(B)}\ \frac 12 \qquad \textbf {(C)}\ \frac{\cos^2 \t
    13 KB (1,948 words) - 12:26, 1 April 2022
  • real r = 5/dir(54).x, h = 5 tan(54*pi/180);
    13 KB (1,966 words) - 09:33, 16 August 2020
  • ...of <math>\tan \angle CBE</math>, <math>\tan \angle DBE</math>, and <math>\tan \angle ABE</math> form a [[geometric progression]], and the values of <math
    13 KB (2,049 words) - 13:03, 19 February 2020
  • ...>L_2</math> and the x-axis, so <math>m=\tan{2\theta}=\frac{2\tan\theta}{1-\tan^2{\theta}}=\frac{120}{119}</math>. We also know that <math>L_1</math> and <
    2 KB (253 words) - 22:52, 29 December 2021
  • ...e positive x- axis, the answer is <math>\dfrac{\tan(BOJ) + \tan(BOJ+60) + \tan(BOJ-60)}{2}</math>. Using <math>\tan(BOJ) = 2</math>, and the tangent addition formula, this simplifies to <math
    2 KB (350 words) - 15:12, 15 July 2018
  • ...BG</math>). Then <math>\tan \angle EOG = \frac{x}{450}</math>, and <math>\tan \angle FOG = \frac{y}{450}</math>. ...frac{y}{450}}{1 - \frac{x}{450} \cdot \frac{y}{450}}.</cmath> Since <math>\tan 45 = 1</math>, this simplifies to <math>1 - \frac{xy}{450^2} = \frac{x + y}
    13 KB (2,052 words) - 15:26, 7 June 2022
  • ...y find that <math>\tan \angle OF_1T=\sqrt{69}/10</math>. Therefore, <math>\tan\angle XOT</math>, which is the desired slope, must also be <math>\sqrt{69}/ ...rac{\sqrt3\cdot\sin\theta}{2\cos\theta}=\frac65\sec\theta-\frac{\sqrt3}{2}\tan\theta</math>
    12 KB (2,000 words) - 13:17, 28 December 2020
  • ...5}</math>. Therefore, <math>\overline{AG} = \frac{52}{5}</math>, so <math>\tan{(\alpha)} = \frac{6}{13}</math>. Our goal now is to use tangent <math>\angl ...}</math> or <math>\frac{126}{137}</math>. Now we solve the equation <math>\tan{\angle EAG} = \frac{126}{137} = \frac{\frac{60-4x}{5}}{\frac{3x+25}{5}}</ma
    13 KB (2,129 words) - 02:46, 31 October 2021
  • ...B'EF=\theta</math>, so <math>\angle B'EA = \pi-2\theta</math>. Then <math>\tan(\pi-2\theta)=\frac{15}{8}</math>, or <cmath>\frac{2\tan(\theta)}{\tan^2(\theta)-1}=\frac{15}{8}</cmath> using supplementary and double angle iden
    8 KB (1,321 words) - 18:58, 13 February 2021
  • ...tan x+\tan y=25</math> and <math>\cot x + \cot y=30</math>, what is <math>\tan(x+y)</math>?
    5 KB (847 words) - 19:20, 24 June 2022
  • In triangle <math>ABC</math>, <math>\tan \angle CAB = 22/7</math>, and the altitude from <math>A</math> divides <mat
    6 KB (902 words) - 08:57, 19 June 2021
  • Suppose that <math>\sec x+\tan x=\frac{22}7</math> and that <math>\csc x+\cot x=\frac mn,</math> where <ma draw(Circle(((1/cos(pi/12))*cos(a*pi/12), (1/cos(pi/12))*sin(a*pi/12)), tan(pi/12)));
    7 KB (1,106 words) - 22:05, 7 June 2021
  • Find the smallest positive integer solution to <math>\tan{19x^{\circ}}=\dfrac{\cos{96^{\circ}}+\sin{96^{\circ}}}{\cos{96^{\circ}}-\si
    6 KB (931 words) - 17:49, 21 December 2018
  • Given that <math>\sum_{k=1}^{35}\sin 5k=\tan \frac mn,</math> where angles are measured in degrees, and <math>m_{}</math
    7 KB (1,094 words) - 13:39, 16 August 2020
  • ...an{(\angle OAB)}=\frac{2-\frac{1}{3}}{1+2\cdot \frac{1}{3}}</cmath><cmath>\tan{(\angle OAB)}=1</cmath><cmath>\cos{(\angle OAB)}=\frac{1}{\sqrt{2}}</cmath>
    8 KB (1,425 words) - 15:08, 26 April 2022
  • .../math>, we have <math>OM = \sqrt{OB^2 - BM^2} =4</math>. This gives <math>\tan \angle BOM = \frac{BM}{OM} = \frac 3 4</math>. ...efore, since <math>\angle AOM</math> is clearly acute, we see that <cmath>\tan \angle AOM =\frac{\sqrt{1 - \cos^2 \angle AOM}}{\cos \angle AOM} = \frac{\s
    19 KB (3,221 words) - 02:42, 3 April 2022

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