# Search results

• ...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 $\tan\theta=2$, [itex]\sin\theta=\frac{2}{\sqrt{5}}, \cos\theta=\frac{1}{\s
2 KB (323 words) - 09:56, 16 September 2022

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