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  • ...y the addition formula, <math>\tan(x+y) = \frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}</math>. Let <math>a = \cot^{-1}(3)</math>, <math>b=\cot^{-1}(7)</math>, ...an(a)=\frac{1}{3},\quad\tan(b)=\frac{1}{7},\quad\tan(c)=\frac{1}{13},\quad\tan(d)=\frac{1}{21}</math>,</p></center>
    3 KB (473 words) - 12:06, 18 December 2018
  • ...ective medians; in other words, <math>\tan \theta_2 = 1</math>, and <math>\tan \theta_2 =2</math>. ...ta_2 - \theta_1) = \frac{\tan \theta_2 - \tan \theta_1}{1 + \tan \theta_1 \tan \theta_2} = \frac{2-1}{1 + 2 \cdot 1 } = \frac{1}{3}. </cmath>
    11 KB (1,722 words) - 17:09, 9 April 2022
  • ...tan x+\tan y=25</math> and <math>\cot x + \cot y=30</math>, what is <math>\tan(x+y)</math>? Since <math>\cot</math> is the reciprocal function of <math>\tan</math>:
    3 KB (527 words) - 10:27, 31 July 2021
  • Let <math>\tan\angle ABC = x</math>. Now using the 1st square, <math>AC=21(1+x)</math> and ...ving, we get <math>\sin{2\theta} = \frac{1}{10}</math>. Now to find <math>\tan{\theta}</math>, we find <math>\cos{2\theta}</math> using the Pythagorean
    5 KB (838 words) - 18:05, 19 February 2022
  • In [[triangle]] <math>ABC</math>, <math>\tan \angle CAB = 22/7</math>, and the [[altitude]] from <math>A</math> divides ...CD = 3</math>. Then <math>\tan \angle DAB = \frac{17}{h}</math> and <math>\tan \angle CAD = \frac{3}{h}</math>. Using the [[Trigonometric_identities#Angle
    1 KB (190 words) - 19:20, 27 February 2018
  • ...\beta)^2-\tan \alpha \tan \beta}{\tan^2 \alpha + 2\tan \alpha \tan \beta +\tan^2 \beta}</math> ...sqrt{995}</math>. We see that <math>\tan \beta = \infty</math>, and <math>\tan \alpha = \sqrt{994}</math>.
    6 KB (961 words) - 20:43, 9 April 2022
  • Let <math>a_{i} = (2i - 1) \tan{\theta_{i}}</math> for <math>1 \le i \le n</math> and <math>0 \le \theta_{i ...that that <math>S_{n} + 17 = \sum_{k = 1}^{n}(2k - 1)(\sec{\theta_{k}} + \tan{\theta_{k}})</math>.
    3 KB (497 words) - 01:43, 25 September 2020
  • draw(Circle(((1/cos(pi/12))*cos(a*pi/12), (1/cos(pi/12))*sin(a*pi/12)), tan(pi/12))); ...h>OA</math> and <math>m \angle MOA = 15^\circ</math>. Thus <math>AM = (1) \tan{15^\circ} = 2 - \sqrt {3}</math>, which is the radius of one of the circles
    4 KB (729 words) - 04:43, 6 December 2019
  • Suppose that <math>\sec x+\tan x=\frac{22}7</math> and that <math>\csc x+\cot x=\frac mn,</math> where <ma ...s#Pythagorean Identities|trigonometric Pythagorean identities]] <math>1 + \tan^2 x = \sec^2 x</math> and <math>1 + \cot^2 x = \csc^2 x</math>.
    8 KB (1,342 words) - 05:20, 21 July 2022
  • Since <math>PC=100</math>, <math>PX=200</math>. So, <math>\tan(\angle OXP)=\frac{OP}{PX}=\frac{50}{200}=\frac{1}{4}</math>. Thus, <math>\tan(\angle BXA)=\tan(2\angle OXP)=\frac{2\tan(\angle OXP)}{1- \tan^2(\angle OXP)} = \frac{2\cdot \left(\frac{1}{4}\right)}{1-\left(\frac{1}{4}
    8 KB (1,243 words) - 00:26, 19 June 2022
  • ...le sum identity gives <cmath>\tan 3x=\tan(2x+x)=\frac{3\tan x-\tan^3x}{1-3\tan^2x}.</cmath> Thus, <math>\frac{3-\tan^2x}{1-3\tan^2x}=11</math>. Solving, we get <math>\tan x= \frac 12</math>. Hence, <math>CM=\frac{11}2</math> and <math>AC= \frac{1
    6 KB (900 words) - 19:54, 4 December 2021
  • Find the smallest positive integer solution to <math>\tan{19x^{\circ}}=\dfrac{\cos{96^{\circ}}+\sin{96^{\circ}}}{\cos{96^{\circ}}-\si ...2\sin{141^{\circ}}\cos{45^{\circ}}}{2\cos{141^{\circ}}\sin{45^{\circ}}} = \tan{141^{\circ}}</math>.
    4 KB (503 words) - 15:46, 3 August 2022
  • \begin{align*}DP&=z\tan\theta\\ EP&=x\tan\theta\\
    6 KB (978 words) - 22:31, 28 May 2021
  • \begin{eqnarray*} \tan \alpha & = & \frac {21}{27} \\ \tan \beta & = & \frac {21}{23} \\
    3 KB (472 words) - 15:59, 25 February 2022
  • Given that <math>\sum_{k=1}^{35}\sin 5k=\tan \frac mn,</math> where angles are measured in degrees, and <math>m_{}</math ...ath>, we get <cmath>s = \frac{1 - \cos 175}{\sin 175} \Longrightarrow s = \tan \frac{175}{2},</cmath> and our answer is <math>\boxed{177}</math>.
    2 KB (322 words) - 11:22, 12 October 2020
  • ...rrow AB=\sqrt{OB^2-AO^2}=\sqrt{5^2-1.4^2}=\frac{24}{5}</math>. Then <math>\tan(\angle ABO)=\frac{OA}{AB}=\frac{7}{24}</math>, so the [[slope]] of line <ma
    3 KB (571 words) - 00:38, 13 March 2014
  • Note that the slope of <math>\overline{AC}</math> is <math>\tan 60^\circ = \sqrt {3}.</math> Hence, the equation of the line containing <ma
    5 KB (769 words) - 20:19, 11 March 2022
  • <cmath>2 > \tan 2x \Longrightarrow x < \frac 12 \arctan 2.</cmath>
    2 KB (284 words) - 13:42, 10 October 2020
  • pair A=(0,0), B=(2,0), C=(1,Tan(37)), M=IP(A--(2Cos(30),2Sin(30)),B--B+(-2,2Tan(23))); pair A=(0,0), B=(2,0), C=(1,Tan(37)), M=IP(A--(2Cos(30),2Sin(30)),B--B+(-2,2Tan(23)));
    7 KB (1,048 words) - 06:16, 20 August 2020
  • Hence <math>x=25\sin\theta=50\cos\theta</math>. Solving <math>\tan\theta=2</math>, <math>\sin\theta=\frac{2}{\sqrt{5}}, \cos\theta=\frac{1}{\s
    2 KB (327 words) - 17:37, 30 July 2022

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