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  • ...ments <math>\overline{AO}, \overline{BO}</math> will have slope <math>\pm \tan{60^{\circ}} = \pm \sqrt{3}</math>. [[Without loss of generality]] consider
    1 KB (184 words) - 13:10, 27 August 2020
  • ...||\cos||<math>\textstyle \sin</math>||\sin||<math>\textstyle \tan</math>||\tan
    12 KB (1,894 words) - 21:24, 7 June 2024
  • (a) <math>\tan(45^\circ)</math> (g) <math>\tan(21\pi)</math>
    2 KB (338 words) - 13:28, 24 March 2020
  • ...Also note that <cmath>AB = 1 = \overline{AA'} + \overline{A'B} = \frac{x}{\tan(15)} + x</cmath> Using the fact <math>\tan(15) = 2-\sqrt{3}</math>, this yields <cmath>x = \frac{1}{3+\sqrt{3}} = \fra
    7 KB (1,067 words) - 12:23, 8 April 2024
  • <math>\tan (-A)=-\tan A</math> <math>\tan (\pi+A) = \tan A</math>
    6 KB (1,097 words) - 18:18, 21 January 2016
  • <math>b \tan{\frac{\omega}{2}} \le c < b</math>
    3 KB (425 words) - 21:18, 20 August 2020
  • ...e CGF</math>. Thus, <math>\angle EGF=\angle FCG</math> and <math>\tan EGF=\tan FCG=\frac{1}{2}</math>. Solving <math>EF=\frac{1}{2}</math>. Adding, the an
    5 KB (738 words) - 13:11, 27 March 2023
  • E = (0,Tan(15)); F = (1 - Tan(15),1);
    4 KB (710 words) - 02:47, 18 April 2024
  • ...al number such that <math>\sec x - \tan x = 2</math>. Then <math>\sec x + \tan x =</math>
    13 KB (1,945 words) - 18:28, 19 June 2023
  • <cmath>\tan\left(\frac{\theta}{2}\right) = \frac{1}{x} = \frac{\sqrt{2}}{4}</cmath> ...3</math> and <math>V_1 = \frac{\pi a^2 \times H_1 H_2}{3} = \frac{\pi a^3 \tan (\angle A_1 A H_1) }{3}</math> .
    7 KB (1,214 words) - 18:49, 29 January 2018
  • Consider the points <math>M_k = (1, \tan k^\circ)</math> in the coordinate plane with origin <math>O=(0,0)</math>, f ...hen the left hand side of the equation simplifies to <math>\tan 89-\tan 0=\tan 89=\frac{\sin 89}{\cos 89}=\frac{\cos 1}{\sin 1}</math> as desired.
    4 KB (628 words) - 07:41, 19 July 2016
  • <math>\text {(A)}\ \sec^2 \theta - \tan \theta \qquad \text {(B)}\ \frac 12 \qquad \text {(C)}\ \frac{\cos^2 \theta ...<cmath> \frac{OB}{OC} = \frac{AB}{AC} \Longrightarrow AC \sec \theta = OC \tan \theta </cmath> We multiply both sides by <math>\cos \theta</math> to simpl
    6 KB (979 words) - 12:50, 17 July 2022
  • 21. Construct <math>sin C, cos C, tan C</math> given unit segment <math>1</math> and acute angle <math>C</math>.
    3 KB (443 words) - 20:52, 28 August 2014
  • ..., </math> <math>\tan, \; \sin^{-1}, \; \cos^{-1}, \,</math> and <math>\, \tan^{-1} \,</math> buttons. The display initially shows 0. Given any positive
    3 KB (540 words) - 13:31, 4 July 2013
  • ...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 ...a)\tan(DBE + \alpha) = \frac {\tan^2 DBE - \tan^2 \alpha}{1 - \tan ^2 DBE \tan^2 \alpha},
    2 KB (302 words) - 19:59, 3 July 2013
  • ..., </math> <math>\tan, \; \sin^{-1}, \; \cos^{-1}, \,</math> and <math>\, \tan^{-1} \,</math> buttons. The display initially shows 0. Given any positive <cmath> \cos \tan^{-1} \sqrt{(n-m)/m} = \sqrt{m/n} . </cmath>
    3 KB (516 words) - 00:18, 6 April 2020
  • ...midpoint of <math>BC</math>. What is the largest possible value of <math>\tan{\angle BAD}</math>?
    13 KB (2,025 words) - 13:56, 2 February 2021
  • ...dpoint]] of <math>BC</math>. What is the largest possible value of <math>\tan{\angle BAD}</math>? ..., and since <math>\tan\angle BAF = \frac{2\sqrt{3}}{x-2}</math> and <math>\tan\angle DAE = \frac{\sqrt{3}}{x-1}</math>, we have
    3 KB (513 words) - 14:35, 7 June 2018
  • ...a</math> with the x-axis and passes through the origin has equation <math>\tan(\theta)x</math>, so the line through <math>A_0</math> and <math>B_1</math>
    9 KB (1,482 words) - 13:52, 4 April 2024
  • Since we are dealing with acute angles, <math>\tan(\arctan{a}) = a</math>. Note that <math>\tan(\arctan{a} + \arctan{b}) = \dfrac{a + b}{1 - ab}</math>, by tangent additio
    3 KB (490 words) - 22:36, 28 November 2023

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