# 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
59 KB (7,185 words) - 17:56, 2 July 2022
• $\textbf {(A)}\ \sec^2 \theta - \tan \theta \qquad \textbf {(B)}\ \frac 12 \qquad \textbf {(C)}\ \frac{\cos^2 \t 13 KB (1,948 words) - 12:26, 1 April 2022 • real r = 5/dir(54).x, h = 5 tan(54*pi/180); 13 KB (1,966 words) - 09:33, 16 August 2020 • ...of [itex]\tan \angle CBE$, $\tan \angle DBE$, 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 $m_{}</math 7 KB (1,094 words) - 13:39, 16 August 2020 • ...an{(\angle OAB)}=\frac{2-\frac{1}{3}}{1+2\cdot \frac{1}{3}}</cmath><cmath>\tan{(\angle OAB)}=1</cmath><cmath>\cos{(\angle OAB)}=\frac{1}{\sqrt{2}}</cmath> 8 KB (1,425 words) - 15:08, 26 April 2022 • .../math>, we have [itex]OM = \sqrt{OB^2 - BM^2} =4$. 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

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