# 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

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