# Search results

• ...we make use of the identity $\tan^2x+1=\sec^2x$. Set $x=a\tan\theta$ and the radical will go away. However, the $dx$ wil Since $\sec^2(\theta)-1=\tan^2(\theta)$, let $x=a\sec\theta$.
1 KB (173 words) - 18:42, 30 May 2021
• * $\tan^2x + 1 = \sec^2x$ * $\tan(x + y) = \frac{\tan (x) + \tan (y)}{1 - \tan (x) \tan (y)}$
8 KB (1,371 words) - 10:33, 7 November 2021
• ...de opposite $A$ to the side adjacent to $A$. <cmath>\tan (A) = \frac{\textrm{opposite}}{\textrm{adjacent}} = \frac{a}{b}.</cmath> ...e reciprocal of the tangent of $A$. <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
• $\textbf {(A)}\ \sec^2 \theta - \tan \theta \qquad \textbf {(B)}\ \frac 12 \qquad \textbf {(C)}\ \frac{\cos^2 \t 13 KB (1,948 words) - 12:26, 1 April 2022 • real r = 5/dir(54).x, h = 5 tan(54*pi/180); 13 KB (1,966 words) - 13:08, 10 July 2022 • ...of [itex]\tan \angle CBE$, $\tan \angle DBE$, and $\tan \angle ABE$ form a [[geometric progression]], and the values of <math
13 KB (2,049 words) - 13:03, 19 February 2020
• ...>L_2[/itex] and the x-axis, so $m=\tan{2\theta}=\frac{2\tan\theta}{1-\tan^2{\theta}}=\frac{120}{119}$. We also know that $L_1$ and <
2 KB (253 words) - 22:52, 29 December 2021
• ...e positive x- axis, the answer is $\dfrac{\tan(BOJ) + \tan(BOJ+60) + \tan(BOJ-60)}{2}$. Using $\tan(BOJ) = 2$, and the tangent addition formula, this simplifies to <math
2 KB (350 words) - 15:12, 15 July 2018
• ...BG[/itex]). Then $\tan \angle EOG = \frac{x}{450}$, and $\tan \angle FOG = \frac{y}{450}$. ...frac{y}{450}}{1 - \frac{x}{450} \cdot \frac{y}{450}}.</cmath> Since $\tan 45 = 1$, this simplifies to $1 - \frac{xy}{450^2} = \frac{x + y} 13 KB (2,052 words) - 15:26, 7 June 2022 • ...y find that [itex]\tan \angle OF_1T=\sqrt{69}/10$. Therefore, $\tan\angle XOT$, which is the desired slope, must also be $\sqrt{69}/ ...rac{\sqrt3\cdot\sin\theta}{2\cos\theta}=\frac65\sec\theta-\frac{\sqrt3}{2}\tan\theta$
12 KB (2,000 words) - 13:17, 28 December 2020
• ...5}[/itex]. Therefore, $\overline{AG} = \frac{52}{5}$, so $\tan{(\alpha)} = \frac{6}{13}$. Our goal now is to use tangent $\angl ...}$ or $\frac{126}{137}$. Now we solve the equation $\tan{\angle EAG} = \frac{126}{137} = \frac{\frac{60-4x}{5}}{\frac{3x+25}{5}}</ma 13 KB (2,129 words) - 02:46, 31 October 2021 • ...B'EF=\theta$, so $\angle B'EA = \pi-2\theta$. Then $\tan(\pi-2\theta)=\frac{15}{8}$, 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[/itex] and $\cot x + \cot y=30$, what is $\tan(x+y)$?
5 KB (847 words) - 19:20, 24 June 2022
• In triangle $ABC$, $\tan \angle CAB = 22/7$, and the altitude from $A$ divides <mat
6 KB (902 words) - 08:57, 19 June 2021
• Suppose that $\sec x+\tan x=\frac{22}7$ and that $\csc x+\cot x=\frac mn,$ 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 $\tan{19x^{\circ}}=\dfrac{\cos{96^{\circ}}+\sin{96^{\circ}}}{\cos{96^{\circ}}-\si 6 KB (931 words) - 17:49, 21 December 2018 • Given that [itex]\sum_{k=1}^{35}\sin 5k=\tan \frac mn,$ where angles are measured in degrees, and $m_{}</math 7 KB (1,094 words) - 13:39, 16 August 2020 • ...an{(\angle OAB)}=\frac{2-\frac{1}{3}}{1+2\cdot \frac{1}{3}}</cmath><cmath>\tan{(\angle OAB)}=1</cmath><cmath>\cos{(\angle OAB)}=\frac{1}{\sqrt{2}}</cmath> 10 KB (1,663 words) - 01:17, 17 July 2022 • .../math>, we have [itex]OM = \sqrt{OB^2 - BM^2} =4$. This gives $\tan \angle BOM = \frac{BM}{OM} = \frac 3 4$. ...efore, since $\angle AOM$ 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, $\tan(x+y) = \frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}$. Let $a = \cot^{-1}(3)$, $b=\cot^{-1}(7)$, ...an(a)=\frac{1}{3},\quad\tan(b)=\frac{1}{7},\quad\tan(c)=\frac{1}{13},\quad\tan(d)=\frac{1}{21}[/itex],</p></center>
3 KB (473 words) - 12:06, 18 December 2018
• ...ective medians; in other words, $\tan \theta_2 = 1$, and $\tan \theta_2 =2$. ...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[/itex] and $\cot x + \cot y=30$, what is $\tan(x+y)$? Since $\cot$ is the reciprocal function of $\tan$:
3 KB (527 words) - 10:27, 31 July 2021
• Let $\tan\angle ABC = x$. Now using the 1st square, $AC=21(1+x)$ and ...ving, we get $\sin{2\theta} = \frac{1}{10}$. Now to find $\tan{\theta}$, we find $\cos{2\theta}$ using the Pythagorean
5 KB (838 words) - 18:05, 19 February 2022
• In [[triangle]] $ABC$, $\tan \angle CAB = 22/7$, and the [[altitude]] from $A$ divides ...CD = 3[/itex]. Then $\tan \angle DAB = \frac{17}{h}$ and $\tan \angle CAD = \frac{3}{h}$. 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}[/itex] ...sqrt{995}[/itex]. We see that $\tan \beta = \infty$, and $\tan \alpha = \sqrt{994}$.
6 KB (961 words) - 20:43, 9 April 2022
• Let $a_{i} = (2i - 1) \tan{\theta_{i}}$ for $1 \le i \le n$ and $0 \le \theta_{i ...that that [itex]S_{n} + 17 = \sum_{k = 1}^{n}(2k - 1)(\sec{\theta_{k}} + \tan{\theta_{k}})$.
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[/itex] and $m \angle MOA = 15^\circ$. Thus $AM = (1) \tan{15^\circ} = 2 - \sqrt {3}$, which is the radius of one of the circles
4 KB (729 words) - 04:43, 6 December 2019
• Suppose that $\sec x+\tan x=\frac{22}7$ and that $\csc x+\cot x=\frac mn,$ where <ma ...s#Pythagorean Identities|trigonometric Pythagorean identities]] $1 + \tan^2 x = \sec^2 x$ and $1 + \cot^2 x = \csc^2 x$.
8 KB (1,342 words) - 05:20, 21 July 2022
• Since $PC=100$, $PX=200$. So, $\tan(\angle OXP)=\frac{OP}{PX}=\frac{50}{200}=\frac{1}{4}$. Thus, $\tan(\angle BXA)=\tan(2\angle OXP)=\frac{2\tan(\angle OXP)}{1- \tan^2(\angle OXP)} = \frac{2\cdot \left(\frac{1}{4}\right)}{1-\left(\frac{1}{4} 8 KB (1,243 words) - 00:26, 19 June 2022 • ...le sum identity gives <cmath>\tan 3x=\tan(2x+x)=\frac{3\tan x-\tan^3x}{1-3\tan^2x}.</cmath> Thus, [itex]\frac{3-\tan^2x}{1-3\tan^2x}=11$. Solving, we get $\tan x= \frac 12$. Hence, $CM=\frac{11}2$ and $AC= \frac{1 6 KB (900 words) - 19:54, 4 December 2021 • Find the smallest positive integer solution to [itex]\tan{19x^{\circ}}=\dfrac{\cos{96^{\circ}}+\sin{96^{\circ}}}{\cos{96^{\circ}}-\si ...2\sin{141^{\circ}}\cos{45^{\circ}}}{2\cos{141^{\circ}}\sin{45^{\circ}}} = \tan{141^{\circ}}$.
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 $\sum_{k=1}^{35}\sin 5k=\tan \frac mn,$ where angles are measured in degrees, and $m_{}</math ...ath>, we get <cmath>s = \frac{1 - \cos 175}{\sin 175} \Longrightarrow s = \tan \frac{175}{2},</cmath> and our answer is [itex]\boxed{177}$.
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}[/itex]. Then $\tan(\angle ABO)=\frac{OA}{AB}=\frac{7}{24}$, so the [[slope]] of line <ma
3 KB (571 words) - 00:38, 13 March 2014
• Note that the slope of $\overline{AC}$ is $\tan 60^\circ = \sqrt {3}.$ 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 $x=25\sin\theta=50\cos\theta$. Solving $\tan\theta=2$, $\sin\theta=\frac{2}{\sqrt{5}}, \cos\theta=\frac{1}{\s 2 KB (327 words) - 17:37, 30 July 2022 • ...we have that [itex]\frac{y}{x}=\tan{\frac{\theta}{2}}$. Let $\tan{\frac{\theta}{2}}=m_1$, for convenience. Therefore if $(x,y)</ma <cmath>\tan{\frac{\theta}{2}}=\sqrt{\frac{1-\cos{\theta}}{1+\cos{\theta}}}</cmath> 7 KB (1,182 words) - 09:56, 7 February 2022 • We have that [itex]\tan(\angle AMO)=\frac{19}{x},$ 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
• ...[/itex] 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 $\tan(A+B)$ 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 $AB$. Let $f(m,n)$ denote the maximum value $\tan^{2}\angle AMP$ for fixed $m$ and $n$ where <mat $\tan{\angle{OAB}} = \dfrac{OT}{AT} = \dfrac{r}{m}$
3 KB (542 words) - 14:05, 4 December 2021
• ...f $AB$. Let $f(m,n)$ denote the maximum value $\tan^{2}\angle AMP$ for fixed $m$ and $n$ where <mat
8 KB (1,355 words) - 14:54, 21 August 2020
• ..., $\frac{AY}{CY}=\sqrt 3,$ and $CY=CX-BX$. If $\tan \angle APB= -\frac{a+b\sqrt{c}}{d},$ where $a,b,$ and <mat ...angle DPB)=270^\circ[/itex], 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 $\tan 15^\circ \tan 25^\circ \tan 35^\circ =\tan \theta$ and $0^\circ \le \theta \le 180^\circ,$ 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 $\tan 15^\circ \tan 25^\circ \tan 35^\circ =\tan \theta$ and $0^\circ \le \theta \le 180^\circ,$ find <mat ...c{BX}{CX}=\frac23[/itex] and $\frac{AY}{CY}=\sqrt 3.$ If $\tan \angle APB= \frac{a+b\sqrt{c}}{d},$ where $a,b,$ and <math
5 KB (848 words) - 23:49, 25 February 2017
• ...>. Thus, $\frac{a}{b} = \tan 15^\circ$ and $\frac{a}{b} = \tan 75^\circ$, and so one of the angles of the triangle must be $15^ 7 KB (1,134 words) - 21:42, 23 June 2021 • | [itex]\frac d{dx} \tan x = \sec^2 x$ | $\frac d{dx} \sec x = \sec x \tan x$
3 KB (506 words) - 16:23, 11 March 2022
• *$\int\tan x\,dx = \ln |\cos x| + C$ *$\int \sec x\,dx = \ln |\sec x + \tan x| + C$
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}[/itex] $\frac{r}{q} = \tan (A/2) \tan (B/2)$.
2 KB (380 words) - 22:12, 19 May 2015
• ...}{4}[/itex] and $\tan{y}=\frac{1}{6}$, find the value of $\tan{x}$.
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, $s$, the length of the apothem is $\frac{s}{2\tan\left(\frac{\pi}{n}\right)}$.
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) $y = \frac{1}{x^2-5x}$ 2) $\tan 3x$. 2) Since $\tan 3x = \frac{\sin 3x}{\cos 3x}$, we need to find where $\cos 3x = 4 KB (664 words) - 11:44, 8 May 2020 • The value of [itex]\tan\left(\Omega\right)$ can be expressed as $\frac{m}{n}$, whe
7 KB (1,135 words) - 23:53, 24 March 2019
• The value of $\tan\left(\Omega\right)$ can be expressed as $\frac{m}{n}$, whe ...ric substitution; namely, define $\theta$ such that $x = \tan{\theta}$. 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}[/itex], and $\text{slope}$$(QA)=\tan{\angle QAB}=\frac{c}{b}\cdot\frac{n+1}{n-1}$.
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}}[/itex]). We see that $\frac rx = \tan \frac{180 - \theta}{2} = \sqrt{\frac{1 - \cos (180 - \theta)}{1 + \cos (180 ...We see that [itex]\frac rx = \tan \left(\frac{180 - 2\theta}{2}\right) = \tan (90 - \theta)$. In terms of $r$, we find that $x = \f 11 KB (1,851 words) - 12:31, 21 December 2021 • ...th>[\triangle EFB'] = \frac{1}{2} (FB' \cdot EF) = \frac{1}{2} (FB') (FB' \tan 75^{\circ})$. 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}[/itex] according to half angle formula. Similarly, we can find $tan\angle NCK=\frac{1}{2}$. So we can see that $JK=ON=14-\frac{7x}{2 8 KB (1,421 words) - 21:58, 31 July 2022 • [itex]b \tan{\frac{\omega}{2}} \le c < b$ ...we require $AX \geqslant AC > AB$. But $\frac{AB}{AX} = \tan{\frac{\omega}{2}}$, so we get the condition in the question
1 KB (205 words) - 04:12, 7 June 2021
• ...||\cos||$\textstyle \sin$||\sin||$\textstyle \tan$||\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 $\tan(15) = 2-\sqrt{3}$, this yields <cmath>x = \frac{1}{3+\sqrt{3}} = \fra
6 KB (914 words) - 17:37, 5 January 2022
• $b \tan{\frac{\omega}{2}} \le c < b$
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 $\sec x - \tan x = 2$. Then $\sec x + \tan x =$
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[/itex] and $V_1 = \frac{\pi a^2 \times H_1 H_2}{3} = \frac{\pi a^3 \tan (\angle A_1 A H_1) }{3}$ .
7 KB (1,214 words) - 18:49, 29 January 2018
• Consider the points $M_k = (1, \tan k^\circ)$ in the coordinate plane with origin $O=(0,0)$, f ...hen the left hand side of the equation simplifies to $\tan 89-\tan 0=\tan 89=\frac{\sin 89}{\cos 89}=\frac{\cos 1}{\sin 1}$ as desired.
4 KB (628 words) - 07:41, 19 July 2016
• $\text {(A)}\ \sec^2 \theta - \tan \theta \qquad \text {(B)}\ \frac 12 \qquad \text {(C)}\ \frac{\cos^2 \theta ...<cmath> \frac{OB}{OC} = \frac{AB}{AC} \Longrightarrow AC \sec \theta = OC \tan \theta </cmath> We multiply both sides by [itex]\cos \theta$ to simpl
6 KB (979 words) - 12:50, 17 July 2022
• 21. Construct $sin C, cos C, tan C$ given unit segment $1$ and acute angle $C$.
3 KB (443 words) - 20:52, 28 August 2014
• ..., [/itex] $\tan, \; \sin^{-1}, \; \cos^{-1}, \,$ and $\, \tan^{-1} \,$ buttons. The display initially shows 0. Given any positive
3 KB (540 words) - 13:31, 4 July 2013
• ...of $\tan \angle CBE$, $\tan \angle DBE$, and $\tan \angle ABE$ 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
• ..., [/itex] $\tan, \; \sin^{-1}, \; \cos^{-1}, \,$ and $\, \tan^{-1} \,$ 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 $BC$. What is the largest possible value of $\tan{\angle BAD}$?
13 KB (2,025 words) - 13:56, 2 February 2021
• ...dpoint]] of $BC$. What is the largest possible value of $\tan{\angle BAD}$? ..., and since $\tan\angle BAF = \frac{2\sqrt{3}}{x-2}$ and $\tan\angle DAE = \frac{\sqrt{3}}{x-1}$, we have
3 KB (513 words) - 14:35, 7 June 2018
• Since we are dealing with acute angles, $\tan(\arctan{a}) = a$. Note that $\tan(\arctan{a} + \arctan{b}) = \dfrac{a + b}{1 - ab}$, 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 $\tan(\theta) = \frac {3}{4}$. Therefore, $XC = \frac {64}{3}$. ...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.[/itex]</center> ...\tan ^2 \theta},\ \tan 3\theta = \frac{3\tan \theta - \tan^3 \theta}{1 - 3\tan^2 \theta}[/itex], and
3 KB (513 words) - 21:46, 12 July 2021
• ...>. Denote $x=\tan{(A/2)}$, $x=\tan{(B/2)}$, $z=\tan{(C/2)}$, 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 $[0,2\pi)$ that satisfy $\tan^2 x - 2\tan x\sin x=0$. Compute $\lfloor10S\rfloor$. Let a and b be the two possible values of $\tan\theta$ given that $\sin\theta + \cos\theta = \dfrac{193}{137}</m 71 KB (11,743 words) - 16:29, 26 November 2021 • ...he other triangles. Thus, the area of triangle [itex]A_1BC=\frac{1}{4}a^2\tan\frac{A}{2}=\frac{1}{4}a^2\left(\frac{2r}{b+c-a}\right)$ 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 $\{\theta_1, \theta_2, \theta_3...\}$ such that $a_n = \tan{\theta_n}$, and $0 \leq \theta_n < 180$. ...+ 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
• ...$\angle ACH$ can be simplified. Indeed, if you know that $\tan(75)=2+\sqrt{3}$ or even take a minute or two to work out the sine and ...= 2 + \sqrt{3}[/itex]. Looking that the answer options we see that $\tan{75^\circ} = 2 + \sqrt{3}$. This means the answer is $D$.
6 KB (988 words) - 13:02, 3 October 2021
• ...}[/itex] and the $x$-axis is $30^{\circ}$, and $tan(30) = \frac{\sqrt{3}}{3}$.
4 KB (707 words) - 16:36, 15 February 2021
• ...e BAD = \angle DAC[/itex]. Notice $\tan \theta = BD$ and $\tan 2 \theta = 2$. By the double angle identity, <cmath>2 = \frac{2 BD}{1
2 KB (359 words) - 11:33, 2 July 2021
• Since we have $\tan OAB = \frac {35}{24}$ and $\tan OBA = \frac{6}{35}$ , we have $\sin {(OAB + OBA)} = \frac {1369} ...ot \tan (\alpha+\beta) = r\cdot \frac{\tan\alpha + \tan\beta}{1-\tan\alpha\tan\beta}= \frac{37^2\cdot r}{18\cdot 35}</cmath> 10 KB (1,657 words) - 22:02, 12 January 2022 • ...BOP$ and $COP$, with $BO=CO=7$ and $OP=7 \tan 15=7(2-\sqrt{3})=14-7\sqrt3$. Then, the area of [$\triangle BPC< 6 KB (1,046 words) - 22:54, 5 June 2022 • <cmath>\frac{NV}{MV} = \frac{\sin (\alpha)}{\sin (90^\circ - \alpha)} = \tan (\alpha)</cmath> ...math>VW = NW + MV - 1 = \frac{1}{1+\frac{3}{4}\cot(\alpha)} + \frac{1}{1+\tan (\alpha)} - 1$. Taking the derivative of $VW$ 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 $\tan{\left(a_0\right)}\tan{\left(a_1\right)}\cdots \tan{\left(a_n\right)}\ge n^{n + 1}$.
2 KB (309 words) - 09:44, 20 July 2016
• If $y(x) = \tan x$, then $\frac{dy}{dx} = \sec^2 x$. Note that this follow
2 KB (288 words) - 00:53, 26 March 2018
• <cmath>\frac{\sqrt{3}}{2}\cos(x)=\frac{3}{2}\sin(x)\implies \tan(x)=\frac{\sqrt{3}}{3}</cmath>
4 KB (608 words) - 03:46, 17 January 2022
• ...ath> has $a_1=\sin x$, $a_2=\cos x$, and $a_3= \tan x$ for some real number $x$. For what value of $n</ma 12 KB (1,845 words) - 13:00, 19 February 2020 • ...sqrt{159}}{7}$, and $ED^2 + EB^2 = 3050$, and that $\tan m \angle ACB$ can be expressed in the form $\frac{a \sqrt{b}}{c} 7 KB (1,297 words) - 01:29, 25 November 2016 • <cmath>r = AX \tan(A/2) = CY \tan(C/2)</cmath> ...^2}{2\cdot 12\cdot 15} = \frac{200}{30\cdot 12}=\frac{5}{9}$, $\tan(A/2) = \sqrt{\frac{1-\cos A}{1+\cos A}} = \sqrt{\frac{9-5}{9+5}} = \frac{2} 12 KB (1,994 words) - 17:41, 23 October 2021 • Hence, [itex](4 \sin^2(x+y) - 7 \cos^2(x+y)) \leq 0$, yielding $\tan^2(x+y) = \frac{7}{4}$, or $x + y = \pm \arctan{\frac{\sqrt{7}}{2 37 KB (6,217 words) - 17:40, 21 June 2022 • [itex]AY \cos(\delta)\sin(\alpha)\tan(\delta) = AY \sin(\delta)\sin(\alpha)$
13 KB (2,178 words) - 14:14, 11 September 2021
• ...se sides. Prove that if <cmath> a+b=\tan{\frac{\gamma}{2}}(a\tan{\alpha}+b\tan{\beta}) </cmath> the triangle is isosceles.
3 KB (509 words) - 12:39, 29 January 2021
• ...ath> has $a_1=\sin x$, $a_2=\cos x$, and $a_3= \tan x$ for some real number $x$. For what value of $n</ma ...ometric sequence, we have [itex]\cos^2x=\sin x \tan x$. Since $\tan x=\frac{\sin x}{\cos x}$, we can rewrite this as $\cos^3x=\sin^2 2 KB (375 words) - 19:52, 24 June 2022 • | Tan 3 KB (295 words) - 20:01, 4 March 2020 • ...\frac{C}{2}=90^\circ-\angle \frac{A}{2}$, so $\cot\frac{C}{2}=\tan\frac{A}{2}$, and we find that $\frac{r}{x}=\frac{x-5}{r}$. ...rac{B}{2}=\frac{14-x}{r}[/itex], and $\cot\frac{D}{2}=\frac{12-x}{r}=\tan\frac{B}{2}=\frac{r}{14-x}$. Therefore, $r^2=\frac{12-x}{14-x}</m 4 KB (753 words) - 18:58, 2 June 2022 • ...- 12x^3 + 54x^2 - 108x + 405} \cdot \cos (\theta) = 36</cmath> and <cmath>\tan (\theta) = \frac{3}{x}</cmath> YAY!!! We have two equations for two variabl Finally, since [itex]\tan (\theta) = \frac{3}{6 - 3 \sqrt{3}} = 2 + \sqrt{3}$, $\theta = \ 4 KB (624 words) - 19:58, 24 October 2021 • <cmath>a = R \tan \frac{B}{2} + R \tan \frac{C}{2} = R \frac{\sin \frac{B+C}{2}}{\cos \frac{B}{2} \cos \frac{C}{2} 7 KB (1,180 words) - 23:13, 8 February 2015 • Suppose the common slope of the lines is [itex]m$ and let $m=\tan\theta$. Then, we want to find <cmath>\cos 2\left(90-\theta\right)=2\c ...rt{2}\right)m^{2}&=4 \\ m^{2}&=\frac{4}{8+8\sqrt{2}} \\ \Rightarrow m^{2}=\tan^{2}\theta=\frac{\sin^{2}\theta}{\cos^{2}\theta}&=\frac{1}{2+2\sqrt{2}}.\end
8 KB (1,344 words) - 22:36, 13 August 2022
• ...eft(a_0-\frac{\pi}{4}\right)+ \tan \left(a_1-\frac{\pi}{4}\right)+\cdots +\tan \left(a_n-\frac{\pi}{4}\right)\geq n-1. </cmath> <cmath> \tan a_0\tan a_1 \cdots \tan a_n\geq n^{n+1}. </cmath>
3 KB (486 words) - 06:11, 24 November 2020
• ...nt to $OB,OC,$ and arc $BC$. It is known that $\tan AOC=\frac{24}{7}$. The ratio $\frac{r_2} {r_1}$ can be exp
8 KB (1,349 words) - 19:10, 14 June 2022
• ...n (\theta + 120)+\tan (\theta-120)}{6}=\frac{3\tan\theta-\tan^3\theta}{2-6\tan^2\theta}</cmath> and <cmath>\begin{align*} ...{3}&=\frac{\tan\theta (\tan(\theta-120)+\tan(\theta+120))+\tan(\theta-120)\tan(\theta+120)}{12}\\
15 KB (2,593 words) - 13:37, 29 January 2021
• \sin, \cos, \tan represents $\sin, \cos, \tan$
2 KB (315 words) - 21:13, 28 February 2022
• ...number such that $\sec x - \tan x = 2$. Then $\sec x + \tan x =$ ...(\sec x + \tan x) = \sec^{2} x - \tan^{2} x = 1[/itex], so $\sec x + \tan x = \boxed{\textbf{(E)}\ 0.5}$.
544 bytes (81 words) - 11:41, 2 January 2016
• ...}) \qquad \mathrm{(D) \ }\tan{15^\circ} \qquad \mathrm{(E) \ } \frac{1}{4}\tan{60^\circ} [/itex]
13 KB (1,879 words) - 14:00, 19 February 2020
• $sin^2(x) + tan^2(x) = -cos^2(x) + \frac{1}{sin^2(x) + cos^2(x)}$ $sin^2(x) + cos^2(x) + tan^2(x) = \frac{1}{sin^2(x) + cos^2(x)}$
8 KB (1,351 words) - 20:30, 10 July 2016
• The slope we are looking for is equivalent to $\tan (\theta + 45)$ where $\angle AOX = \theta$. Using tangent <cmath> \tan (\theta + 45)= \frac{\tan \theta + \tan 45}{1-\tan\theta\tan 45} = \frac{\frac{1}{\sqrt{2}}+1}{1-\frac{1}{\sqrt{2}}}=3+2\sqrt{2}</cmath>
3 KB (453 words) - 20:25, 8 March 2021
• ...times\sin{3x}}{\cos{2x}\times\cos{3x}}=1 [/itex], or $\tan{2x}\times\tan{3x}=1$. ...identity]] $-\tan{x}=\tan{-x}$, we have $\tan{2x}\times\tan{-3x}=-1$.
3 KB (493 words) - 18:16, 4 June 2021
• <cmath> a+b=\tan{\frac{\gamma}{2}}(a\tan{\alpha}+b\tan{\beta}) </cmath> ...at as $\gamma = \pi -\alpha-\beta$ then and the identity $\tan\left(\frac \pi 2 - x \right)=\cot x$ our equation becomes:
2 KB (416 words) - 17:54, 13 January 2022
• ...}) \qquad \mathrm{(D) \ }\tan{15^\circ} \qquad \mathrm{(E) \ } \frac{1}{4}\tan{60^\circ} [/itex] ...irc}}{2\cos{15^\circ}\cos{5^\circ}}=\frac{\sin{15^\circ}}{\cos{15^\circ}}=\tan{15^\circ}, \boxed{\text{D}} [/itex].
1 KB (159 words) - 12:52, 5 July 2013
• ...can also use trig manipulation on $BCE$ to get that $CE=a\tan{\beta}$. $[BED]=\frac{BD\cdot CE}{2}=\frac{ac\cos{\beta}\tan{\beta}}{4}=\frac{ac\sin{\beta}}{4}$
2 KB (332 words) - 20:37, 12 February 2017
• ...ath> and $B=25^\circ$, then the value of $(1+\tan A)(1+\tan B)$ is ...\ }2 \qquad \mathrm{(C) \ } 1+\sqrt{2} \qquad \mathrm{(D) \ } 2(\tan A+\tan B) \qquad \mathrm{(E) \ }\text{none of these} [/itex]
17 KB (2,445 words) - 14:00, 19 February 2020
• ...ath> and $B=25^\circ$, then the value of $(1+\tan A)(1+\tan B)$ is ...\ }2 \qquad \mathrm{(C) \ } 1+\sqrt{2} \qquad \mathrm{(D) \ } 2(\tan A+\tan B) \qquad \mathrm{(E) \ }\text{none of these} [/itex]
4 KB (718 words) - 21:49, 31 July 2020
• ...counter-clockwise order and right angle at $A$, let $f(t)=\tan(\angle{CBA})$. What is <cmath>\prod_{t\in T} f(t)?</cmath> ...on $A'B'C'$ labeled that way will give us $\tan CBA \cdot \tan C'B'A' = 1$. First we consider the reflection about the line $y= 2 KB (356 words) - 17:10, 4 April 2020 • Now we have [itex] BE=BA\cdot\tan\angle EAB=1\cdot\tan30^\circ=\frac{\sqrt{3}}{3}$. Finally, $[A 2 KB (364 words) - 16:28, 30 December 2011 • ...nt to [itex]OB,OC,$ and arc $BC$. It is known that $\tan AOC=\frac{24}{7}$. The ratio $\frac{r_2} {r_1}$ can be exp ...h>, so $\cos AOC=\frac{7}{25}$. Now, we have $\tan FOD=\tan\frac{AOC}{2}=\sqrt{\frac{1-\cos AOC}{1+\cos AOC}}=\sqrt{\frac{1-\frac{7}{25 3 KB (432 words) - 14:12, 2 January 2012 • ...t triangles [itex]\Delta PO_1H, \Delta O_1S_1O_2$. Similarly, $\tan{\angle PO_2O_1}=\frac{h}{b}=\frac{52}{39}$ using right triangles <mat
3 KB (522 words) - 21:25, 3 January 2012
• $\cot 10+\tan 5 =$ Find the sum of the roots of $\tan^2x-9\tan x+1=0$ that are between $x=0$ and $x=2\pi$ radi
15 KB (2,247 words) - 13:44, 19 February 2020
738 bytes (126 words) - 21:56, 17 October 2016
• Find the sum of the roots of $\tan^2x-9\tan x+1=0$ that are between $x=0$ and $x=2\pi$ radi ...[/itex] are positive and distinct, so by considering the graph of $y=\tan x$, the smallest two roots of the original equation $x_1,\ x_2</ 2 KB (282 words) - 21:14, 2 March 2019 • [itex]\cot 10+\tan 5=$ We have <cmath>\cot 10 +\tan 5=\frac{\cos 10}{\sin 10}+\frac{\sin 5}{\cos 5}=\frac{\cos10\cos5+\sin10\si
534 bytes (69 words) - 16:11, 25 February 2022
• ...counter-clockwise order and right angle at $A$, let $f(t)=\tan(\angle{CBA})$. What is <cmath>\prod_{t\in T} f(t)?</cmath>
19 KB (2,648 words) - 19:56, 10 August 2020
• ...[/itex] so that $\cos\theta=\sin(90-\theta)=s/6$. Then $m=\tan\theta=3$. Substituting into $\left(\frac{4m^2+10}{m^2+1},\frac{6 ...$. Setting $6\sin\theta=-2\cos\theta$, we get that $\tan\theta=-1/3$. This means $-1/3$ is the slope of line $11 KB (1,881 words) - 19:54, 10 August 2020 • ...\angle C.$ If $\frac{DE}{BE} = \frac{8}{15},$ then $\tan B$ can be written as $\frac{m \sqrt{p}}{n},$ where $m 10 KB (1,617 words) - 12:16, 13 March 2020 • ...\angle C.$ If $\frac{DE}{BE} = \frac{8}{15},$ then $\tan B$ can be written as $\frac{m \sqrt{p}}{n},$ where $m ...[itex]FD = 4a\sqrt{3}$, and $FB = 11a$. Finally, $\tan{B} = \tfrac{DF}{FB}=\tfrac{4\sqrt{3}a}{11a} = \tfrac{4\sqrt{3}}{11}$.
9 KB (1,513 words) - 16:43, 31 December 2021
• pair tan = reflect(origin,(5,y))*(0,y); draw((5,y)--tan,linetype("4 4"));
3 KB (478 words) - 03:06, 5 April 2012
• Evaluate: $\int(x\tan^{-1}x)dx$ <cmath> \frac{d}{dx}\tanh^{-1}\tan x </cmath>
3 KB (525 words) - 13:59, 27 May 2012
• ...exist two positive numbers $x$ such that $\sin(\arccos(\tan(\arcsin(x))))=x$. Find the product of the two possible $x </ma 6 KB (910 words) - 17:32, 27 May 2012 • ..., Q=(\cot x, \cot^2 x), R=(\sin x, \sin^2 x)$ and $S =(\tan x, \tan^2 x)$ are the vertices of a trapezoid. What is $\sin(2x)$?
16 KB (2,459 words) - 02:46, 30 January 2021
• ..., Q=(\cot x, \cot^2 x), R=(\sin x, \sin^2 x)[/itex] and $S =(\tan x, \tan^2 x)$ are the vertices of a trapezoid. What is $\sin(2x)$? Let $f,g,h,j$ be $\sin, \cos, \tan, \cot$ (not respectively). Then we have four points $(f,f^2),(g, 2 KB (375 words) - 00:54, 28 September 2021 • ...} \cdot \frac{r}{s-b} \cdot \frac{r}{s-c} = \frac{1}{4} \tan A/2 \tan B/2 \tan C/2.</cmath> Lemma. [itex]\tan x \tan (A - x)$ is increasing on $0 < x < \frac{A}{2}$, where <ma
2 KB (376 words) - 23:29, 18 May 2015
• ...tter. Only the numerator, because we are trying to find $\frac{P}{Q}=\tan\text{arg}(\Sigma)$ a PROPORTION of values. So denominators would canc
10 KB (1,631 words) - 21:39, 9 March 2022
16 KB (2,451 words) - 04:27, 6 September 2021
• real r = 5/dir(54).x, h = 5 tan(54*pi/180);
1 KB (237 words) - 23:06, 3 February 2020
• $\tanh(x)= -1\tan{iz}$
423 bytes (78 words) - 23:33, 22 May 2013
• <cmath>\frac{XC}{CY}=\tan {\angle CYZ}=\tan (90-\alpha)</cmath> <cmath>\frac{CQ}{CY}=\tan {\angle CYQ}=\tan (\alpha+\beta).</cmath>
7 KB (1,250 words) - 18:05, 1 October 2021
• $\tan{\frac{3A}{2}}\tan{\frac{3B}{2}}=1$ Note that $\tan{x}=\frac{1}{\tan(90-x)}$, or $\tan{x}\tan(90-x)=1$
5 KB (861 words) - 10:00, 3 May 2021
• ...an{\angle{C}}-1)m}{i\tan{\angle{C}}}.</cmath> We wish to simplify $(i\tan{\angle{C}}-1)m$ first. Note that <cmath>m=\frac{|CM|}{|CA|}\cdot(a)=\ (i\tan{\angle{C}}-1)m&=(i\tan{\angle{C}}-1)((|BC|)\cos{\angle{C}}(\cos{\angle{C}}+i\sin{\angle{C}}))\\
11 KB (1,983 words) - 15:48, 6 June 2022
• ...ow EC=20-10 \sqrt 3[/itex]. (It is important to memorize the sin, cos, and tan values of $15^\circ$ and $75^\circ$.) Therefore, we h
10 KB (1,609 words) - 05:43, 20 June 2022
• ...an angle $\theta$ relative to the coordinate axis, where $\tan\theta = \tfrac 34$. We rotate the coordinate axis by angle $\the 4 KB (662 words) - 18:46, 22 November 2021 • ...tan {\theta})^4 - 3(\tan {\theta})^2 + 1 = 0.$ This gives us $(\tan {\theta})^2 = \dfrac{3+\sqrt{5}}{2}\longrightarrow \boxed{E}$
4 KB (696 words) - 07:02, 15 August 2021
• ...\left(\tfrac{w-z}{z}\right) [/itex]. The maximum possible value of $\tan^2 \theta$ can be written as $\tfrac{p}{q}$, where $p< 9 KB (1,472 words) - 13:59, 30 November 2021 • ...\left(\tfrac{w-z}{z}\right)$. The maximum possible value of $\tan^2 \theta$ can be written as $\tfrac{p}{q}$, where $p< We know that [itex]\tan{\theta}$ is equal to the imaginary part of the above expression divid
5 KB (749 words) - 13:15, 21 August 2021
• ...nd $\sin \frac12 \theta = \sqrt{\frac{x-1}{2x}}$, then $\tan \theta$ equals
18 KB (2,788 words) - 13:55, 20 February 2020
• ..._2)}{1-\tan^2(\theta_2)} = 4\tan(\theta_2)[/itex]. Solving, we have $\tan(\theta_2) = 0, \dfrac{\sqrt{2}}{2}$. But line $L_1$ is not
2 KB (237 words) - 18:02, 20 March 2018
• ...c})+\log_{10}(\tan 3^{\circ})+\cdots+\log_{10}(\tan 88^{\circ})+\log_{10}(\tan 89^{\circ}). [/itex]
16 KB (2,291 words) - 13:45, 19 February 2020
• $\textbf{(A)}\ \tan \theta = \theta\qquad \textbf{(B)}\ \tan \theta = 2\theta\qquad 17 KB (2,512 words) - 14:00, 19 February 2020 • ...angle at [itex]C$. If $\sin A = \frac{2}{3}$, then $\tan B$ is If $\tan{\alpha}$ and $\tan{\beta}$ are the roots of $x^2 - px + q = 0$, and $\co 15 KB (2,309 words) - 23:43, 2 December 2021 • If [itex]\sin x+\cos x=1/5$ and $0\le x<\pi$, then $\tan x$ is
15 KB (2,437 words) - 04:39, 26 November 2021
• If $\tan x=\dfrac{2ab}{a^2-b^2}$ where $a>b>0$ and $0^\circ <x real x = 6-h*tan(t); 17 KB (2,732 words) - 13:54, 20 February 2020 • ...}{4}$ and $\tan{y}=\frac{1}{6}$, find the value of $\tan{x}$. $\tan(x+\arctan\frac{1}{6})=\tan\frac{\pi}{4}=1$
1 KB (209 words) - 16:57, 10 June 2018
• ...PK = \dfrac{1}{2}a \tan \dfrac{1}{2}C[/itex] and $QL = \dfrac{1}{2}b \tan \dfrac{1}{2}C$ from right triangles $\triangle PKC$ and <m <cmath>= \dfrac{\frac{1}{2}a\tan\frac{1}{2}C \cdot (a + b)}{a\sin\frac{1}{2}C} </cmath>
8 KB (1,480 words) - 14:52, 5 August 2022
• ...ation by $\cos{(x)}$ to get <cmath>\frac{\sin{(x)}}{\cos{(x)}}=\tan{(x)}=3.</cmath>
2 KB (402 words) - 20:53, 24 August 2021
• ...c})+\log_{10}(\tan 3^{\circ})+\cdots+\log_{10}(\tan 88^{\circ})+\log_{10}(\tan 89^{\circ}). [/itex] ...b}[/itex], the answer is $\log_{10} {\tan 1^\circ \tan 2^\circ \dots \tan 89^\circ} = \log_{10} 1 = 0.$ $\boxed{\textbf{(A)}}.$
1 KB (164 words) - 12:42, 31 March 2018
• $\textbf{(A)}\ \tan \theta = \theta\qquad \textbf{(B)}\ \tan \theta = 2\theta\qquad 2 KB (301 words) - 18:50, 1 April 2018 • ...}{DP}, DP = \frac{1}{\tan 54}</cmath>Therefore, [itex]AB = 2DP = \frac{2}{\tan 54}$. ...efore, $AO + AQ + AR = AO + 2AQ = \frac{1}{\sin 54}+\frac{4 \sin 72}{\tan 54} = \frac{1}{\sin 54} + 8 \sin 36 \cos 54 = \frac{1}{\cos 36} + 8-8\cos^2 4 KB (702 words) - 17:13, 17 April 2020 • ..., we get<cmath>\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}.</cmath> 2 KB (458 words) - 19:24, 2 February 2020 • ...> is an isosceles [itex]30 - 75 - 75$ triangle. Thus, $DF = CF \tan 15^\circ = \dfrac{5}{\sqrt{2}} (2 - \sqrt{3})$ by the Half-Angle form
4 KB (729 words) - 16:33, 6 December 2020
• Setting the two areas equal, we get <cmath>\tan A = \frac{\sin A}{\cos A} = 8 \iff \sin A = \frac{8}{\sqrt{65}}, \cos A = \ <cmath>\tan{\angle{CYE}} = \frac{1}{8}</cmath>
25 KB (4,164 words) - 21:22, 7 July 2022
• Let angle $\angle XAB=A$, which is an acute angle, $\tan{A}=t$, then $X=(1-a,at)$.
5 KB (902 words) - 09:58, 20 August 2021
• Let angle $\angle XAB=A$, which is an acute angle, $\tan{A}=t$, then $X=(1-a,at)$.
4 KB (760 words) - 16:45, 29 April 2020
• From Alice's point of view, $\tan(\theta)=\frac{z}{y}$. $\tan{30}=\frac{\sin{30}}{\cos{30}}=\frac{1}{\sqrt{3}}$. So, $y=z*\sqr From Bob's point of view, [itex]\tan(\theta)=\frac{z}{x}$. $\tan{60}=\frac{\sin{60}}{\cos{60}}=\sqrt{3}$. So, $x = \frac{z}{\sqrt 3 KB (492 words) - 22:26, 14 October 2020 • <cmath>\frac{31}{40} \geq \tan\theta</cmath> However, [itex]\tan\theta = \tan(\frac{90-A}{2}) = \frac{\sin(90-A)}{\cos(90-A)+1} = \frac{\cos A}{\sin A + 9 KB (1,526 words) - 02:31, 29 December 2021 • ...and [itex]KQ=y$, assuming WLOG $x>y$, we must have $\tan(120^{\circ})=-\sqrt{3}=\dfrac{\dfrac{x+y}{100 \sqrt{3}}}{\dfrac{30000-xy}{3 ...[itex]\sqrt{3}\tan{\left(\alpha\right)}$, we can set $\sqrt{3}\tan{\left(\alpha\right)}=a$ for convenience since we really only care abo
14 KB (2,321 words) - 10:59, 14 July 2022
• If $\tan{\alpha}$ and $\tan{\beta}$ are the roots of $x^2 - px + q = 0$, and $\co ...cot\theta=\frac{1}{\tan\theta}$, we have $\frac{1}{\tan(\alpha)\tan(\beta)}=\frac{1}{q}=s$.
1 KB (222 words) - 00:58, 20 February 2019
• ...s clear that $I = \left(\frac{b + c – a}{2} , \frac{b + c – a}{2}\tan(A / 2)\right)$. ...and the $y$ coordinate of $O$ is $-\frac{b}{2} \tan{B-90}$. From this, $(5)$ follows in this case as well.
6 KB (1,078 words) - 13:35, 26 July 2016
• ...angle at $C$. If $\sin A = \frac{2}{3}$, then $\tan B$ is so $\tan{B} = \frac{x \sqrt{5}}{2x} = \frac{\sqrt{5}}{2}$, which is choice <ma
1 KB (171 words) - 00:42, 20 February 2019
• ...{2}+\frac{1}{3}}{1-(\frac{1}{2})(\frac{1}{3})}[/itex]. Simplifying, $\tan(\theta_a + \theta_b) = 1$, so $\theta_a + \theta_b$ in rad
2 KB (363 words) - 12:46, 10 May 2022
• Let $f(x) = \sin{x} + 2\cos{x} + 3\tan{x}$, using radian measure for the variable $x$. In what in ...tan function. Upon further examination, it is clear that the positive the tan function creates will balance the other two functions, and thus the first s
3 KB (564 words) - 14:12, 23 October 2021
• Let $f(x) = \sin{x} + 2\cos{x} + 3\tan{x}$, using radian measure for the variable $x$. In what in
15 KB (2,418 words) - 14:43, 12 August 2020
• Then $\tan \alpha = \frac {2\sqrt{3}}{5},$ <cmath>x(\frac {\sin \beta}{\tan{\alpha}} - \cos \beta) +x (\frac {\cos\beta}{2} +\frac{\sin\beta \sqrt{3}}{
18 KB (2,941 words) - 12:12, 17 July 2022
• | 67 || Senpai-Tan || 80 || 8151.479 || 101.893
187 KB (10,824 words) - 18:27, 3 February 2022
• We start by letting $\tan x = \frac{\sin x}{\cos x}$ so that our equation is now: <cmath>\frac{
616 bytes (107 words) - 22:01, 25 December 2017
• real x = 6-h*tan(t); real y = x*tan(2*t);
3 KB (431 words) - 19:52, 23 June 2021
• How many solutions does the equation $\tan{(2x)} = \cos{(\tfrac{x}{2})}$ have on the interval $[0, 2\pi]?</ 14 KB (2,073 words) - 15:15, 21 October 2021 • If [itex]\tan a$ and $\tan b$ are the roots of $x^2+px+q=0$, then compute, in terms o
2 KB (377 words) - 14:52, 7 January 2018
• &\tan(2b)= &\frac{1}{4}\\&
571 bytes (90 words) - 05:19, 17 June 2021
• $\sin x\left(1+\tan x\tan\frac{x}{2}\right)=4-\cot x$
7 KB (1,127 words) - 18:23, 11 January 2018
• ...\frac{2\sin(46)\cos(10)}{-2\sin(46)\sin({-10})}=\frac{\sin(80)}{\cos(80)}=\tan(80)</cmath>
12 KB (1,878 words) - 22:11, 23 October 2021
• ...c{7\pi}{6}[/itex] without the loss of generality. Since $\tan(2\phi)>\tan\frac{\pi}{3},$ we deduce that $2\phi>\frac{\pi}{3},$ from
10 KB (1,662 words) - 12:45, 13 September 2021
• <cmath>\frac {(r-r_1)\cdot (r-r_2)}{r_1 \cdot r_2} =\tan\beta \tan\gamma.</cmath> <cmath>1 -\frac{2r}{h} = \frac {b+c-a}{b+c+a} = \frac {r}{r_a} = \tan\beta \tan\gamma .</cmath>
13 KB (2,143 words) - 12:30, 16 July 2022
• ...nd $\sin \frac12 \theta = \sqrt{\frac{x-1}{2x}}$, then $\tan \theta$ equals <cmath>\tan \frac{\theta}{2} = \sqrt{\frac{x-1}{2x}} \div \sqrt{\frac{x+1}{2x}}</cmath>
1 KB (184 words) - 14:00, 20 February 2020
• ...}{2}\right )\right )=\tan \left (\frac{1}{2} \right )[/itex]. Since $\tan \left(\frac{\theta}{2} \right ) = \frac{1-\cos \left(\theta \right )}{\sin 1 KB (243 words) - 01:10, 20 February 2019 • ...ath> in the interval [itex][0,2\pi)$ that satisfy $\tan^2 x - 2\tan x\sin x=0$. Compute $\lfloor10S\rfloor$. ...0[/itex]. By the Zero Product Property, $\tan x = 0$ or $\tan x = 2\sin x$.
969 bytes (158 words) - 19:00, 12 July 2018
• Let a and b be the two possible values of $\tan\theta$ given that $\sin\theta + \cos\theta = \dfrac{193}{137}</m ...the sum formula for tangent, the sum of the two possible values of [itex]\tan \theta$ is
2 KB (343 words) - 20:35, 4 August 2018
• ...all triangles $ABC$ which have property: $\tan A,\tan B,\tan C$ are positive integers. Prove that all triangles in $S$
3 KB (439 words) - 12:39, 4 September 2018
• \tan(2x) &= \frac{\sqrt{3}}{4}.
3 KB (558 words) - 20:13, 4 January 2019
• \tan(\angle AIS + \angle DIS) &= -\tan(\angle BIT + \angle CIT) \\ ...ngle DIS} &= \frac{\tan \angle BIT + \tan \angle CIT}{1 - \tan \angle BIT \tan \angle CIT}
3 KB (586 words) - 23:47, 8 January 2019
• ...th> such that for nonnegative integers $n$, the value of $\tan{\left(2^{n}\theta\right)}$ is positive when $n$ is a multi ...90^{\circ} \pmod{180^{\circ}}[/itex]. Also notice that the only way $\tan{\left(2^{n}\theta\right)}$ can be positive for all $n$ tha
6 KB (972 words) - 16:04, 14 March 2021
• <cmath>FG = AG - FG = \frac{r_a}{\tan \left( \frac{A}{2} \right)} - b \cos (A)</cmath> <cmath>FH = AH - AF = \frac{r_a}{\tan \left( \frac{A}{2} \right)} - c \cos (A)</cmath>
10 KB (1,536 words) - 20:27, 12 April 2021
• ...ouble Angle Identity yields $\tan 2\theta = \frac34$, so $\tan (90 - 2\theta) = \frac43$.
4 KB (722 words) - 20:53, 27 March 2019
• ...65}{28}=\frac{28\sin\alpha}{8\cos\alpha}=\frac{7}{2}\tan\alpha\Rightarrow \tan\alpha=\frac{65}{98}</cmath> Thus, $MK=\frac{MN}{\tan\alpha}=98$, so $MO=MK-KO=\boxed{090}$.
8 KB (1,298 words) - 18:08, 14 March 2021
• ...th> such that for nonnegative integers $n,$ the value of $\tan(2^n\theta)$ is positive when $n$ is a multiple of $3< 7 KB (1,254 words) - 12:11, 4 January 2021 • <cmath>\tan a = \frac{8}{15}</cmath> <cmath>A = \frac{1}{2}* 9*\frac{9}{2}\tan a = \frac{54}{5}</cmath> 6 KB (1,030 words) - 18:08, 26 December 2020 • x = tan-1 ( 4 / 3 ) = 0.927 (to 3 decimals) 3 KB (543 words) - 15:24, 13 June 2019 • ...lpha) + \tan^{-1}(\beta) + \tan^{-1} (\gamma).</cmath> The value of [itex]\tan(\omega)$ can be written as $\tfrac{m}{n}$ where $m</m 6 KB (1,052 words) - 13:52, 9 June 2020 • Evaluate: [itex] \int(x\tan^{-1}x)dx$ \int(x\tan^{-1}x)dx &= \frac{x^2}{2}\tan^{-1}x-\int\frac{x^2}{2(x^2+1)}dx\\
670 bytes (116 words) - 18:31, 14 January 2020
• Using the identity that $\tan(x) = -\tan(-x)$
1 KB (175 words) - 18:30, 14 January 2020
• ...all triangles $ABC$ which have property: $\tan A,\tan B,\tan C$ are positive integers. Prove that all triangles in $S$ ...B+C = 180^\circ[/itex], so $z = \tan C = \tan (180^\circ - (A+B)) = -\tan(A+B)$.
3 KB (465 words) - 12:00, 26 September 2019
• We compute that $\cos{\angle{ABC}}=\frac{1}{8}$, so $\tan{\angle{ABC}}=3\sqrt{7}$. ...n \angle ABC}{1 + \cos \angle ABC} = \frac{\sqrt{7}}{3}[/itex], and $\tan \angle DAF = \frac{\sqrt{7}}{7}$.
35 KB (5,206 words) - 18:09, 17 June 2022
• <cmath> \sin^3{x}(1+\cot{x})+\cos^3{x}(1+\tan{x})=\cos{2x} </cmath> ...= 0[/itex], we can divide both sides by $\cos{x}$ to get $\tan{x} = -1$. Thus, $x = \frac{3 \pi}{4} + \pi n$, where <mat
2 KB (305 words) - 12:51, 6 December 2019
• ...opposite angle to $x$ be $\theta$, and let $t:=\tan\frac{\theta}{2}$; let the [[area]] be $A$ and the [[semipe
4 KB (674 words) - 16:03, 25 February 2021
• <cmath> \sin^3{x}(1+\cot{x})+\cos^3{x}(1+\tan{x})=\cos{2x} </cmath>
2 KB (393 words) - 13:39, 4 December 2019
• ...t value of $x$ $(0 < x < \frac{\pi}{2})$ does $\tan x + \cot x$ achieve its minimum?
3 KB (413 words) - 13:10, 21 January 2020
• ...-\beta) = 0[/itex], $\tan(\beta) = \frac{1}{2000}$, find $\tan(\alpha)$.
4 KB (618 words) - 13:33, 21 January 2020
• ...t can be used to demonstrate trigonometric functions such as sin, cos, and tan, but it is most commonly used to visualize the complex numbers. This is don
741 bytes (131 words) - 11:50, 22 January 2020
• ...lve the quadratic, taking the positive solution (C is acute) to get $\tan{C} = \frac{1}{3}.$ So if $AB = a,$ then $BC = 3a</mat ...n angle bisector of [itex]\triangle ABC$ (because we will get $\tan(x) = 1$).
14 KB (2,144 words) - 19:21, 15 May 2022
• How many solutions does the equation $\tan(2x)=\cos(\tfrac{x}{2})$ have on the interval $[0,2\pi]?$ We count the intersections of the graphs of $y=\tan(2x)$ and $y=\cos\left(\frac x2\right):$
4 KB (615 words) - 04:07, 8 July 2022
• ...\frac{3}{5}[/itex], and the angle we are rotating around is x, then $\tan x = \frac{3}{5}$ $\tan(x+45^{\circ}) = \frac{\tan x + \tan(45^{\circ})}{1-\tan x*\tan(45^{\circ})} = \frac{0.6+1}{1-0.6} = \frac{1.6}{0.4} = 4$
6 KB (966 words) - 13:27, 16 June 2022
• ...mula area = apothem * perimeter / 2, the area of the octagon is then 1/2 * tan 67.5 * 8 / 2. Area(quadrilateral)/Area(hexagon)=4*sin^2(67.5) / (2*tan(67.5)) = 2*sin^2(67.5) / (sin(67.5)/cos(67.5)) = 2*sin(67.5)*cos(67.5) = si
10 KB (1,519 words) - 10:45, 19 April 2021
• ==Solution 4 (tan)== ...45^{\circ}+\theta)=\frac{\tan(45^{\circ})+\tan(\theta)}{1-\tan(45^{\circ})\tan(\theta)} = \frac{1+a}{1-a}[/itex]. Since the slope of one line is $6</ 5 KB (881 words) - 01:04, 29 January 2021 • A = (0, tan(3 * pi / 7)); 4 KB (638 words) - 19:41, 17 June 2022 • How many solutions does the trigonometric equation [itex]tan(cos(x)) = cos((x\pi^{2} - sin(x))$ have in the interval $[-\pi, 9 KB (1,450 words) - 18:33, 21 April 2020 • ...[itex]\tan \left(\frac{\angle B}{2}\right) = \frac{r}{2}$ and $\tan \left(\frac{\angle E}{2}\right) = \frac{r}{4}$, so <cmath>\cos (\angle B) =\frac{1-\tan^2 \left(\frac{\angle B}{2}\right)}{1+\tan^2 \left(\frac{\angle B}{2}\right)} = \frac{4-r^2}{4+r^2}</cmath>and
12 KB (2,056 words) - 04:37, 18 June 2022
• ...ngles from the larger rectangle, we get  Area = $33-3BG=33-\frac{9}{\tan(\angle DAE)}$. $\alpha=\tan^{-1}\left(\frac{3}{11}\right)$
8 KB (1,281 words) - 02:36, 20 December 2021
• The angle $\theta$ between diagonals satisfies <cmath>\tan{\frac{\theta}{2}}=\sqrt{\frac{(s-b)(s-d)}{(s-a)(s-c)}}</cmath> (see https:/ ...\tan{\frac{\theta}{2}}=\sqrt{\frac{(11-4)(11-6)}{(11-5)(11-7)}}\text{ or }\tan{\frac{\theta}{2}}=\sqrt{\frac{(11-5)(11-7)}{(11-4)(11-6)}}.</cmath>
15 KB (2,434 words) - 14:10, 21 June 2022
• ...lpha) + \tan^{-1}(\beta) + \tan^{-1} (\gamma).</cmath> The value of $\tan(\omega)$ can be written as $\tfrac{m}{n}$ where $m</m ...ma - \gamma\alpha}$. Substituting Vieta's formulas, we obtain $\tan(\omega) = \frac{\frac{799}{4} - \frac{1}{4}}{1 - (-50)} = \frac{\frac{798}{ 1 KB (192 words) - 18:03, 13 September 2020 • Hence, [itex]\tan \theta = \frac{1}{3}$.
16 KB (2,262 words) - 15:34, 5 July 2022
• ...the acute angle formed by the diagonals of the quadrilateral. Then $\tan \theta$ can be written in the form $\tfrac{m}{n}$, where <
8 KB (1,342 words) - 01:55, 14 July 2021
• Let $\tan A=a.$ The area of the rectangle created by the four equations can be
10 KB (1,662 words) - 09:38, 28 December 2021
• If $\sin x+\cos x=1/5$ and $0\le x<\pi$, then $\tan x$ is Since $\tan x = \frac{\sin x}{\cos x}$, we have
1 KB (207 words) - 21:10, 13 February 2021
• ...the acute angle formed by the diagonals of the quadrilateral. Then $\tan \theta$ can be written in the form $\tfrac{m}{n}$, where < ...and $\cos \theta$ separately and use their values to get $\tan \theta$. We can start by drawing a diagram. Let the vertices of the q
10 KB (1,618 words) - 15:57, 21 June 2022
• <cmath>\tan \alpha =\frac { \sin 2 \alpha}{1+\cos 2 \alpha} = \frac {4/5}{1 - 3/5}=2.</ The radius of the inscribed circle is $r = (s – B'D) \tan \alpha.$
13 KB (1,963 words) - 13:44, 18 June 2022
• ...^\circ[/itex], so $\angle{EAB} = 30^{\circ}$. Then $EB=AE\tan 30^\circ = \sqrt{3}$; therefore $BC=EC-EB=3-\sqrt{3}$. Thu
4 KB (659 words) - 10:03, 7 June 2022
• ...d $MP\perp AB$. Let $\theta = \angle BAC$; so $\tan\theta = \tfrac 68 = \tfrac 34$. Then <cmath>OP=MP-MO=AM\cot\theta - BM\tan\theta = 5(\tfrac 43 - \tfrac 34) = \boxed{\textbf{(C)}\ \tfrac{35}{12}}.</c
4 KB (660 words) - 18:31, 6 June 2022
• ...erp[/itex] $\overline{AB}$ and $AY = 3$. Thus $\tan(30^\circ) = \frac{OY}{3}$ so $OY = \sqrt{3}$.
4 KB (575 words) - 17:46, 31 July 2022
• ~MathFun1000 (Inspired by Way Tan)
2 KB (429 words) - 21:22, 7 April 2022
• ...could find the length of the apothem by the formula $\frac{s}{2\text{tan}(\frac{180}{n})},$ where $s$ is the side length and $6 KB (1,000 words) - 18:54, 16 August 2022 • ...th the positive x axis. Thus, line [itex]m$ makes an angle of $\tan^{-1}(5) - 45^\circ$ with the positive x axis. Thus, the slope of line <cmath> \tan (\tan^{-1}(5) - 45^\circ) = \frac{5 - 1}{1 + 5\cdot 1} = \frac{2}{3},</cmath>
8 KB (1,246 words) - 19:20, 29 June 2022
• ...th>\theta_{3}[/itex]. Note that $\tan(\theta_{1})=1$ and $\tan(\theta_{3})=3$, and this is why we named them as such. Let the angle \tan(\theta_k-\theta_1)&=\tan(\theta_3-\theta_k)\\
13 KB (2,002 words) - 14:30, 2 July 2022