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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
    60 KB (7,207 words) - 23:18, 4 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) - 13:08, 10 July 2022
  • ...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>
    10 KB (1,663 words) - 01:17, 17 July 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
  • ...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
  • ...we have that <math>\frac{y}{x}=\tan{\frac{\theta}{2}}</math>. Let <math>\tan{\frac{\theta}{2}}=m_1</math>, for convenience. Therefore if <math>(x,y)</ma <cmath>\tan{\frac{\theta}{2}}=\sqrt{\frac{1-\cos{\theta}}{1+\cos{\theta}}}</cmath>
    7 KB (1,182 words) - 09:56, 7 February 2022
  • We have that <math>\tan(\angle AMO)=\frac{19}{x},</math> so <cmath>\tan(\angle M)=\tan (2\cdot \angle AMO)=\frac{38x}{x^{2}-361}.</cmath>
    4 KB (658 words) - 19:15, 19 December 2021
  • ...</math> to get <cmath>\tan(\arctan(\tfrac{19}{r})+\arctan(\tfrac{26}{r}))+\tan(\arctan(\tfrac{37}{r})+\arctan(\tfrac{23}{r}))=n\cdot0=0.</cmath> Use the identity for <math>\tan(A+B)</math> again to get <cmath>\frac{\tfrac{45}{r}}{1-19\cdot\tfrac{26}{r^
    2 KB (399 words) - 12:41, 4 November 2021
  • <cmath> \frac{a-b}{a+b}=\frac{\tan [\frac{1}{2}(A-B)]}{\tan [\frac{1}{2}(A+B)]} . </cmath> ...2\sin s \cos d} = \frac{\tan d}{\tan s} = \frac{\tan [\frac{1}{2} (A-B)]}{\tan[ \frac{1}{2} (A+B)]} </cmath>
    2 KB (261 words) - 17:49, 2 March 2017
  • ...}{\sqrt{1 - \cos^2 (x)}} + \frac{\cos(x)}{\sqrt{1 - \sin^2 (x) }} + \frac{\tan(x)}{\sqrt{\sec^2 (x) - 1}} + \frac{\cot (x)}{\sqrt{\csc^2 (x) - 1}}</cmath>
    14 KB (2,102 words) - 22:03, 26 October 2018
  • ...f <math>AB</math>. Let <math>f(m,n)</math> denote the maximum value <math>\tan^{2}\angle AMP</math> for fixed <math>m</math> and <math>n</math> where <mat <math>\tan{\angle{OAB}} = \dfrac{OT}{AT} = \dfrac{r}{m}</math>
    3 KB (542 words) - 14:05, 4 December 2021
  • ...f <math>AB</math>. Let <math>f(m,n)</math> denote the maximum value <math>\tan^{2}\angle AMP</math> for fixed <math>m</math> and <math>n</math> where <mat
    8 KB (1,355 words) - 14:54, 21 August 2020
  • ..., <math>\frac{AY}{CY}=\sqrt 3,</math> and <math>CY=CX-BX</math>. If <math>\tan \angle APB= -\frac{a+b\sqrt{c}}{d},</math> where <math>a,b,</math> and <mat ...angle DPB)=270^\circ</math>, we have <cmath>\begin{align*}\tan\angle APB&=\tan[270^\circ-(\angle APE+\angle BPD)]\\&=\cot (\angle APE+\angle BPD)\\&=-\dfr
    2 KB (358 words) - 23:22, 3 May 2014
  • 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 ...rc}=\frac{\sin 5^\circ(1+2\cos 20^\circ)}{\cos 5^\circ(1+2\cos 20^\circ)}=\tan 5^\circ</cmath>
    1 KB (157 words) - 10:51, 4 April 2012
  • 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^
    7 KB (1,134 words) - 21:42, 23 June 2021
  • | <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 (5,143 words) - 20:49, 28 December 2021
  • ...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
  • \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 \\
    9 KB (1,327 words) - 20:59, 19 February 2019
  • ...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
    8 KB (1,421 words) - 21:58, 31 July 2022
  • <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
    13 KB (2,050 words) - 20:57, 10 August 2022
  • ...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
    6 KB (914 words) - 17:37, 5 January 2022
  • <math>b \tan{\frac{\omega}{2}} \le c < b</math>
    3 KB (425 words) - 21:18, 20 August 2020
  • E = (0,Tan(15)); F = (1 - Tan(15),1);
    5 KB (825 words) - 13:49, 24 October 2021
  • ...al number such that <math>\sec x - \tan x = 2</math>. Then <math>\sec x + \tan x =</math>
    13 KB (1,945 words) - 13:58, 16 December 2020
  • <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
  • 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
    2 KB (404 words) - 17:59, 18 March 2020
  • <cmath>\begin{align*}\tan{37}\times (1008-x) &= \tan{53} \times x\\ \frac{(1008-x)}{x} &= \frac{\tan{53}}{\tan{37}} = \frac{\sin{53}}{\cos{53}} \times\frac{\sin{37}}{\cos{37}}\end{align*
    8 KB (1,206 words) - 00:31, 2 January 2022
  • ...om the [[trigonometric identity|half-angle identity]], we find that <math>\tan(\theta) = \frac {3}{4}</math>. Therefore, <math>XC = \frac {64}{3}</math>. ...now drop altitude AY to solve for tan2A ; now since we know tan2A we know tan A = r/x in terms of r hence solve the resulting equation in r
    6 KB (1,065 words) - 20:12, 9 August 2022
  • ...tarrow (2-\sqrt{3}k)\cos x\le k\sin x\rightarrow \frac{2-\sqrt{3}k}{k}\le \tan x,</cmath>
    6 KB (1,000 words) - 13:52, 16 August 2020
  • ...c{AC(\tan 3\theta - \tan 2\theta)}{AC \tan 2\theta} = \frac{\tan 3\theta}{\tan 2\theta} - 1.</math></center> ...\tan ^2 \theta},\ \tan 3\theta = \frac{3\tan \theta - \tan^3 \theta}{1 - 3\tan^2 \theta}</math>, and
    3 KB (513 words) - 21:46, 12 July 2021
  • ...>. Denote <math>x=\tan{(A/2)}</math>, <math>x=\tan{(B/2)}</math>, <math>z=\tan{(C/2)}</math>, then we have, <cmath>z = \tan{(C/2)} = \tan{(90- (A+B)/2))} = \frac{1-xy}{x+y} </cmath>
    4 KB (703 words) - 18:40, 3 January 2019
  • ...ath> in the interval <math>[0,2\pi)</math> that satisfy <math>\tan^2 x - 2\tan x\sin x=0</math>. Compute <math>\lfloor10S\rfloor</math>. Let a and b be the two possible values of <math>\tan\theta</math> given that <math>\sin\theta + \cos\theta = \dfrac{193}{137}</m
    71 KB (11,743 words) - 16:29, 26 November 2021
  • ...he other triangles. Thus, the area of triangle <math>A_1BC=\frac{1}{4}a^2\tan\frac{A}{2}=\frac{1}{4}a^2\left(\frac{2r}{b+c-a}\right)</math> and similarly
    3 KB (568 words) - 11:50, 30 January 2021
  • ...\tan\frac{A}{2}\sin B\tan\frac{B}{2}} = 2\sqrt{\sin A\tan\frac{B}{2}\sin B\tan\frac{A}{2}} \\ &\leq \sin A\tan\frac{B}{2} + \sin B\tan\frac{A}{2} \\
    4 KB (799 words) - 18:28, 1 July 2015
  • ...ce <math>\{\theta_1, \theta_2, \theta_3...\}</math> such that <math>a_n = \tan{\theta_n}</math>, and <math>0 \leq \theta_n < 180</math>. ...+ 2}} & = \frac {\tan{\theta_n} + \tan{\theta_{n + 1}}}{1 - \tan{\theta_n}\tan{\theta_{n + 1}}} \\
    7 KB (979 words) - 10:53, 23 December 2020
  • ...<math>\angle ACH</math> can be simplified. Indeed, if you know that <math>\tan(75)=2+\sqrt{3}</math> or even take a minute or two to work out the sine and ...= 2 + \sqrt{3}</math>. Looking that the answer options we see that <math>\tan{75^\circ} = 2 + \sqrt{3}</math>. This means the answer is <math>D</math>.
    6 KB (988 words) - 13:02, 3 October 2021
  • ...}</math> and the <math>x</math>-axis is <math>30^{\circ}</math>, and <math>tan(30) = \frac{\sqrt{3}}{3}</math>.
    4 KB (707 words) - 16:36, 15 February 2021
  • ...e BAD = \angle DAC</math>. Notice <math>\tan \theta = BD</math> and <math>\tan 2 \theta = 2</math>. By the double angle identity, <cmath>2 = \frac{2 BD}{1
    2 KB (359 words) - 11:33, 2 July 2021
  • Since we have <math>\tan OAB = \frac {35}{24}</math> and <math>\tan OBA = \frac{6}{35}</math> , we have <math>\sin {(OAB + OBA)} = \frac {1369} ...ot \tan (\alpha+\beta) = r\cdot \frac{\tan\alpha + \tan\beta}{1-\tan\alpha\tan\beta}= \frac{37^2\cdot r}{18\cdot 35}</cmath>
    10 KB (1,657 words) - 22:02, 12 January 2022
  • ...BOP</math> and <math>COP</math>, with <math>BO=CO=7</math> and <math>OP=7 \tan 15=7(2-\sqrt{3})=14-7\sqrt3</math>. Then, the area of [<math>\triangle BPC<
    6 KB (1,046 words) - 22:54, 5 June 2022
  • <cmath>\frac{NV}{MV} = \frac{\sin (\alpha)}{\sin (90^\circ - \alpha)} = \tan (\alpha)</cmath> ...math>VW = NW + MV - 1 = \frac{1}{1+\frac{3}{4}\cot(\alpha)} + \frac{1}{1+\tan (\alpha)} - 1</math>. Taking the derivative of <math>VW</math> with respect
    11 KB (1,849 words) - 10:26, 9 August 2022
  • \sum_{n = 0}^\infty \frac{E_n}{n!} x^n = \sec x + \tan x .
    2 KB (246 words) - 12:50, 6 August 2009
  • ...rac {\pi}{4}\right)} + \tan{\left(a_1 - \frac {\pi}{4}\right)} + \cdots + \tan{\left(a_n - \frac {\pi}{4}\right)}\ge n - 1</cmath> Prove that <math>\tan{\left(a_0\right)}\tan{\left(a_1\right)}\cdots \tan{\left(a_n\right)}\ge n^{n + 1}</math>.
    2 KB (309 words) - 09:44, 20 July 2016
  • If <math>y(x) = \tan x</math>, then <math>\frac{dy}{dx} = \sec^2 x</math>. Note that this follow
    2 KB (288 words) - 00:53, 26 March 2018

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