# 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 (1,393 words) - 21:27, 27 March 2023
• ...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) - 20:15, 7 September 2023
• | Tan
60 KB (7,280 words) - 21:31, 20 September 2023
• $\textbf {(A)}\ \sec^2 \theta - \tan \theta \qquad \textbf {(B)}\ \frac 12 \qquad \textbf {(C)}\ \frac{\cos^2 \t 13 KB (1,948 words) - 12:26, 1 April 2022 • real r = 5/dir(54).x, h = 5 tan(54*pi/180); 13 KB (1,987 words) - 18:53, 10 December 2022 • ...of [itex]\tan \angle CBE$, $\tan \angle DBE$, 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
4 KB (761 words) - 09:10, 1 August 2023
• ...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 $1 - \frac{xy}{450^2} = \frac{x + y} 13 KB (2,080 words) - 21:20, 11 December 2022 • ...y find that [itex]\tan \angle OF_1T=\sqrt{69}/10$. 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 (1,322 words) - 11:02, 25 June 2023
• ...tan x+\tan y=25[/itex] and $\cot x + \cot y=30$, what is $\tan(x+y)$?
5 KB (847 words) - 15:48, 21 August 2023
• 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 $m_{}</math 7 KB (1,094 words) - 13:39, 16 August 2020 • ...an{(\angle OAB)}=\frac{2-\frac{1}{3}}{1+2\cdot \frac{1}{3}}</cmath><cmath>\tan{(\angle OAB)}=1</cmath><cmath>\cos{(\angle OAB)}=\frac{1}{\sqrt{2}}</cmath> 10 KB (1,723 words) - 16:05, 1 June 2023 • .../math>, we have [itex]OM = \sqrt{OB^2 - BM^2} =4$. 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) - 01:05, 7 February 2023
• ...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, $\tan \theta_1 = 1$, 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) - 09:49, 13 September 2023
• ...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 (962 words) - 14:15, 18 December 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 (740 words) - 19:33, 28 December 2022
• 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 $1 + \cot^2 x = \csc^2 x$.
10 KB (1,590 words) - 14:04, 20 January 2023
• 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 $AC= \frac{1 7 KB (1,181 words) - 13:47, 3 February 2023 • Find the smallest positive integer solution to [itex]\tan{19x^{\circ}}=\dfrac{\cos{96^{\circ}}+\sin{96^{\circ}}}{\cos{96^{\circ}}-\si ...2\sin{141^{\circ}}\cos{45^{\circ}}}{2\cos{141^{\circ}}\sin{45^{\circ}}} = \tan{141^{\circ}}$.
4 KB (503 words) - 15:46, 3 August 2022
• \begin{align*}DP&=z\tan\theta\\ EP&=x\tan\theta\\
7 KB (1,184 words) - 13:25, 22 December 2022
• \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 (342 words) - 18:00, 25 September 2023
• ...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) - 19:15, 20 July 2023
• <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 (1,058 words) - 01:41, 6 December 2022
• Hence $x=25\sin\theta=50\cos\theta$. Solving $\tan\theta=2$, $\sin\theta=\frac{2}{\sqrt{5}}, \cos\theta=\frac{1}{\s 2 KB (323 words) - 09:56, 16 September 2022 • ...we have that [itex]\frac{y}{x}=\tan{\frac{\theta}{2}}$. 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 (306 words) - 16:11, 21 February 2023
• ...}{\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}}[/itex]< | <center>$\tan^2 x + 1 = \sec^2 x$ </center>
2 KB (331 words) - 00:37, 26 January 2023
• ...}{\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

View (previous 50 | next 50) (20 | 50 | 100 | 250 | 500)