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  • <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</math>. Taking the derivative of <math>VW</math> with respect
    11 KB (1,849 words) - 19:43, 2 January 2023
  • \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 <math>\tan{\left(a_0\right)}\tan{\left(a_1\right)}\cdots \tan{\left(a_n\right)}\ge n^{n + 1}</math>.
    2 KB (322 words) - 13:31, 23 August 2023
  • If <math>y(x) = \tan x</math>, then <math>\frac{dy}{dx} = \sec^2 x</math>. 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 (621 words) - 15:12, 21 June 2023
  • ...ath> has <math>a_1=\sin x</math>, <math>a_2=\cos x</math>, and <math>a_3= \tan x</math> for some real number <math>x</math>. For what value of <math>n</ma
    12 KB (1,845 words) - 13:00, 19 February 2020
  • ...sqrt{159}}{7}</math>, and <math>ED^2 + EB^2 = 3050</math>, and that <math>\tan m \angle ACB</math> can be expressed in the form <math>\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}</math>, <math>\tan(A/2) = \sqrt{\frac{1-\cos A}{1+\cos A}} = \sqrt{\frac{9-5}{9+5}} = \frac{2}
    14 KB (2,210 words) - 13:14, 11 January 2024
  • Hence, <math>(4 \sin^2(x+y) - 7 \cos^2(x+y)) \leq 0</math>, yielding <math>\tan^2(x+y) = \frac{7}{4}</math>, or <math>x + y = \pm \arctan{\frac{\sqrt{7}}{2
    36 KB (6,214 words) - 20:22, 13 July 2023
  • <math>AY \cos(\delta)\sin(\alpha)\tan(\delta) = AY \sin(\delta)\sin(\alpha)</math>
    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 <math>a_1=\sin x</math>, <math>a_2=\cos x</math>, and <math>a_3= \tan x</math> for some real number <math>x</math>. For what value of <math>n</ma ...ometric sequence, we have <math>\cos^2x=\sin x \tan x</math>. Since <math>\tan x=\frac{\sin x}{\cos x}</math>, we can rewrite this as <math>\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}</math>, so <math>\cot\frac{C}{2}=\tan\frac{A}{2}</math>, and we find that <math>\frac{r}{x}=\frac{x-5}{r}</math>. ...rac{B}{2}=\frac{14-x}{r}</math>, and <math>\cot\frac{D}{2}=\frac{12-x}{r}=\tan\frac{B}{2}=\frac{r}{14-x}</math>. Therefore, <math>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 <math>\tan (\theta) = \frac{3}{6 - 3 \sqrt{3}} = 2 + \sqrt{3}</math>, <math>\theta = \
    5 KB (782 words) - 23:04, 31 August 2023
  • <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 (1,189 words) - 01:22, 19 November 2023
  • Suppose the common slope of the lines is <math>m</math> and let <math>m=\tan\theta</math>. 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) - 18:39, 9 February 2023
  • <cmath>\textrm{tan }^{2} \alpha = \frac{\frac{9z-16}{9z}}{\frac{16}{9z}} = \frac{9z-16}{16}.< <cmath>\textrm{tan } \alpha = \textrm{tan } \angle MOP = \frac{MP}{OM} = \frac{14-y}{20}.</cmath>
    11 KB (1,720 words) - 03:12, 18 December 2023
  • <cmath>AP = AB \cdot (\sin 45^\circ + \cos 45^\circ \cdot \tan 30^\circ),</cmath>
    13 KB (2,055 words) - 05:25, 9 September 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 <math>OB,OC,</math> and arc <math>BC</math>. It is known that <math>\tan AOC=\frac{24}{7}</math>. The ratio <math>\frac{r_2} {r_1}</math> 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 <math>\sin, \cos, \tan</math>
    2 KB (315 words) - 21:13, 28 February 2022
  • ...number such that <math> \sec x - \tan x = 2</math>. Then <math> \sec x + \tan x =</math> ...(\sec x + \tan x) = \sec^{2} x - \tan^{2} x = 1</math>, so <math>\sec x + \tan x = \boxed{\textbf{(E)}\ 0.5}</math>.
    931 bytes (144 words) - 19:36, 1 May 2023
  • ...}) \qquad \mathrm{(D) \ }\tan{15^\circ} \qquad \mathrm{(E) \ } \frac{1}{4}\tan{60^\circ} </math>
    13 KB (1,879 words) - 14:00, 19 February 2020
  • <math>sin^2(x) + tan^2(x) = -cos^2(x) + \frac{1}{sin^2(x) + cos^2(x)}</math> <math>sin^2(x) + cos^2(x) + tan^2(x) = \frac{1}{sin^2(x) + cos^2(x)}</math>
    8 KB (1,351 words) - 20:30, 10 July 2016
  • The slope we are looking for is equivalent to <math>\tan (\theta + 45)</math> where <math>\angle AOX = \theta</math>. 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>
    4 KB (614 words) - 20:09, 12 September 2022
  • ...times\sin{3x}}{\cos{2x}\times\cos{3x}}=1 </math>, or <math> \tan{2x}\times\tan{3x}=1 </math>. ...identity]] <math> -\tan{x}=\tan{-x} </math>, we have <math> \tan{2x}\times\tan{-3x}=-1 </math>.
    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 <math>\gamma = \pi -\alpha-\beta</math> then and the identity <math>\tan\left(\frac \pi 2 - x \right)=\cot x</math> 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} </math> ...irc}}{2\cos{15^\circ}\cos{5^\circ}}=\frac{\sin{15^\circ}}{\cos{15^\circ}}=\tan{15^\circ}, \boxed{\text{D}} </math>.
    1 KB (159 words) - 12:52, 5 July 2013
  • ...can also use trig manipulation on <math>BCE</math> to get that <math>CE=a\tan{\beta}</math>. <math>[BED]=\frac{BD\cdot CE}{2}=\frac{ac\cos{\beta}\tan{\beta}}{4}=\frac{ac\sin{\beta}}{4}</math>
    2 KB (303 words) - 20:28, 2 October 2023
  • ...= 25 \degree</math>, then the value of <math>\left(1+\tan A\right)\left(1+\tan B\right)</math> is ...\qquad \mathrm{(C) \ } 1+\sqrt{2} \qquad \mathrm{(D) \ } 2\left(\tan A+\tan B\right) \qquad \mathrm{(E) \ }\text{none of these} </math>
    17 KB (2,488 words) - 03:26, 20 March 2024
  • ...= 25 \degree</math>, then the value of <math>\left(1+\tan A\right)\left(1+\tan B\right)</math> is ...\qquad \mathrm{(C) \ } 1+\sqrt{2} \qquad \mathrm{(D) \ } 2\left(\tan A+\tan B\right) \qquad \mathrm{(E) \ }\text{none of these} </math>
    5 KB (904 words) - 22:25, 19 March 2024
  • ...counter-clockwise order and right angle at <math>A</math>, let <math>f(t)=\tan(\angle{CBA})</math>. What is <cmath>\prod_{t\in T} f(t)?</cmath> ...on <math>A'B'C'</math> labeled that way will give us <math>\tan CBA \cdot \tan C'B'A' = 1</math>. First we consider the reflection about the line <math>y=
    2 KB (356 words) - 17:10, 4 April 2020
  • Now we have <math> BE=BA\cdot\tan\angle EAB=1\cdot\tan30^\circ=\frac{\sqrt{3}}{3} </math>. Finally, <math> [A
    2 KB (376 words) - 22:06, 23 December 2022
  • ...nt to <math>OB,OC,</math> and arc <math>BC</math>. It is known that <math>\tan AOC=\frac{24}{7}</math>. The ratio <math>\frac{r_2} {r_1}</math> can be exp ...h>, so <math> \cos AOC=\frac{7}{25} </math>. Now, we have <math> \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 <math>\Delta PO_1H, \Delta O_1S_1O_2</math>. Similarly, <math>\tan{\angle PO_2O_1}=\frac{h}{b}=\frac{52}{39}</math> using right triangles <mat
    3 KB (522 words) - 21:25, 3 January 2012
  • <math> \cot 10+\tan 5 = </math> Find the sum of the roots of <math>\tan^2x-9\tan x+1=0</math> that are between <math>x=0</math> and <math>x=2\pi</math> radi
    15 KB (2,247 words) - 13:44, 19 February 2020
  • \text{(C) } tan^2\theta\quad
    738 bytes (126 words) - 21:56, 17 October 2016
  • Find the sum of the roots of <math>\tan^2x-9\tan x+1=0</math> that are between <math>x=0</math> and <math>x=2\pi</math> radi ...</math> are positive and distinct, so by considering the graph of <math>y=\tan x</math>, the smallest two roots of the original equation <math>x_1,\ x_2</
    2 KB (282 words) - 21:14, 2 March 2019
  • <math>\cot 10+\tan 5=</math> 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 <math>A</math>, let <math>f(t)=\tan(\angle{CBA})</math>. What is <cmath>\prod_{t\in T} f(t)?</cmath>
    20 KB (2,681 words) - 09:47, 29 June 2023
  • ...</math> so that <math>\cos\theta=\sin(90-\theta)=s/6</math>. Then <math>m=\tan\theta=3</math>. Substituting into <math>\left(\frac{4m^2+10}{m^2+1},\frac{6 ...</math>. Setting <math>6\sin\theta=-2\cos\theta</math>, we get that <math>\tan\theta=-1/3</math>. This means <math>-1/3</math> is the slope of line <math>
    12 KB (2,183 words) - 21:05, 23 December 2023
  • ...\angle C.</math> If <math>\frac{DE}{BE} = \frac{8}{15},</math> then <math>\tan B</math> can be written as <math>\frac{m \sqrt{p}}{n},</math> where <math>m
    10 KB (1,617 words) - 14:49, 2 June 2023
  • ...\angle C.</math> If <math>\frac{DE}{BE} = \frac{8}{15},</math> then <math>\tan B</math> can be written as <math>\frac{m \sqrt{p}}{n},</math> where <math>m ...<math>FD = 4a\sqrt{3}</math>, and <math>FB = 11a</math>. Finally, <math>\tan{B} = \tfrac{DF}{FB}=\tfrac{4\sqrt{3}a}{11a} = \tfrac{4\sqrt{3}}{11}</math>.
    9 KB (1,523 words) - 12:23, 7 September 2022
  • 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: <math> \int(x\tan^{-1}x)dx </math> <cmath> \frac{d}{dx}\tanh^{-1}\tan x </cmath>
    3 KB (525 words) - 13:59, 27 May 2012
  • ...exist two positive numbers <math> x </math> such that <math> \sin(\arccos(\tan(\arcsin(x))))=x </math>. Find the product of the two possible <math> x </ma
    6 KB (910 words) - 17:32, 27 May 2012
  • ...rac{2}{3}</math>, <math>\tan{\frac{B}{2}} = \frac{1}{2}</math>, and <math>\tan{\frac{C}{2}} = \frac{4}{x},</math> so we have the equation <math>\frac{1}{2
    2 KB (314 words) - 21:17, 31 December 2023
  • ..., Q=(\cot x, \cot^2 x), R=(\sin x, \sin^2 x)</math> and <math>S =(\tan x, \tan^2 x)</math> are the vertices of a trapezoid. What is <math>\sin(2x)</math>?
    16 KB (2,459 words) - 02:46, 30 January 2021
  • ..., Q=(\cot x, \cot^2 x), R=(\sin x, \sin^2 x)</math> and <math>S =(\tan x, \tan^2 x)</math> are the vertices of a trapezoid. What is <math>\sin(2x)</math>? Let <math>f,g,h,j</math> be <math>\sin, \cos, \tan, \cot</math> (not respectively). Then we have four points <math>(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. <math>\tan x \tan (A - x)</math> is increasing on <math>0 < x < \frac{A}{2}</math>, where <ma
    2 KB (376 words) - 23:29, 18 May 2015
  • ...tter. Only the numerator, because we are trying to find <math>\frac{P}{Q}=\tan\text{arg}(\Sigma)</math> a PROPORTION of values. So denominators would canc
    10 KB (1,641 words) - 20:03, 3 January 2024
  • \textbf{(C) } \tan^2\theta\qquad
    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
  • <math>\tanh(x)= -1\tan{iz}</math>
    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
  • <math>\tan{\frac{3A}{2}}\tan{\frac{3B}{2}}=1</math> Note that <math>\tan{x}=\frac{1}{\tan(90-x)}</math>, or <math>\tan{x}\tan(90-x)=1</math>
    5 KB (875 words) - 17:56, 2 October 2023
  • ...an{\angle{C}}-1)m}{i\tan{\angle{C}}}.</cmath> We wish to simplify <math>(i\tan{\angle{C}}-1)m</math> 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 (1,991 words) - 01:31, 19 November 2023
  • ...ow EC=20-10 \sqrt 3</math>. (It is important to memorize the sin, cos, and tan values of <math>15^\circ</math> and <math>75^\circ</math>.) Therefore, we h
    12 KB (1,821 words) - 18:16, 29 October 2023
  • ...an angle <math>\theta</math> relative to the coordinate axis, where <math>\tan\theta = \tfrac 34</math>. We rotate the coordinate axis by angle <math>\the
    4 KB (661 words) - 16:18, 2 September 2022
  • ...tan {\theta})^4 - 3(\tan {\theta})^2 + 1 = 0.</math> This gives us <math>(\tan {\theta})^2 = \dfrac{3+\sqrt{5}}{2}\longrightarrow \boxed{E}</math>
    4 KB (703 words) - 16:24, 9 September 2022
  • ...\left(\tfrac{w-z}{z}\right) </math>. The maximum possible value of <math>\tan^2 \theta</math> can be written as <math>\tfrac{p}{q}</math>, where <math>p<
    9 KB (1,472 words) - 13:59, 30 November 2021
  • ...\left(\tfrac{w-z}{z}\right) </math>. The maximum possible value of <math>\tan^2 \theta</math> can be written as <math>\tfrac{p}{q}</math>, where <math>p< We know that <math>\tan{\theta}</math> is equal to the imaginary part of the above expression divid
    5 KB (782 words) - 20:25, 10 October 2023
  • ...nd <math> \sin \frac12 \theta = \sqrt{\frac{x-1}{2x}}</math>, then <math> \tan \theta</math> equals
    18 KB (2,788 words) - 13:55, 20 February 2020
  • ..._2)}{1-\tan^2(\theta_2)} = 4\tan(\theta_2)</math>. Solving, we have <math>\tan(\theta_2) = 0, \dfrac{\sqrt{2}}{2}</math>. But line <math>L_1</math> 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}). </math>
    16 KB (2,291 words) - 13:45, 19 February 2020
  • <math>\textbf{(A)}\ \tan \theta = \theta\qquad \textbf{(B)}\ \tan \theta = 2\theta\qquad
    17 KB (2,512 words) - 18:30, 12 October 2023
  • ...angle at <math>C</math>. If <math>\sin A = \frac{2}{3}</math>, then <math>\tan B</math> is If <math>\tan{\alpha}</math> and <math>\tan{\beta}</math> are the roots of <math>x^2 - px + q = 0</math>, and <math>\co
    15 KB (2,309 words) - 23:43, 2 December 2021
  • If <math>\sin x+\cos x=1/5</math> and <math>0\le x<\pi</math>, then <math>\tan x</math> is
    15 KB (2,432 words) - 01:06, 22 February 2024
  • If <math>\tan x=\dfrac{2ab}{a^2-b^2}</math> where <math>a>b>0</math> and <math>0^\circ <x real x = 6-h*tan(t);
    17 KB (2,732 words) - 13:54, 20 February 2020
  • ...}{4}</math> and <math>\tan{y}=\frac{1}{6}</math>, find the value of <math>\tan{x}</math>. <math>\tan(x+\arctan\frac{1}{6})=\tan\frac{\pi}{4}=1</math>
    2 KB (266 words) - 21:30, 4 February 2023
  • ...PK = \dfrac{1}{2}a \tan \dfrac{1}{2}C</math> and <math>QL = \dfrac{1}{2}b \tan \dfrac{1}{2}C</math> from right triangles <math>\triangle PKC</math> 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 <math>\cos{(x)}</math> 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}). </math> ...b}</math>, the answer is <math>\log_{10} {\tan 1^\circ \tan 2^\circ \dots \tan 89^\circ} = \log_{10} 1 = 0.</math> <math>\boxed{\textbf{(A)}}.</math>
    1 KB (164 words) - 12:42, 31 March 2018
  • <math>\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, <math>AB = 2DP = \frac{2}{\tan 54}</math>. ...efore, <math>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 <math>30 - 75 - 75</math> triangle. Thus, <math>DF = CF \tan 15^\circ = \dfrac{5}{\sqrt{2}} (2 - \sqrt{3})</math> by the Half-Angle form
    9 KB (1,513 words) - 19:38, 12 November 2023
  • 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>
    31 KB (5,086 words) - 19:15, 20 December 2023
  • Let angle <math>\angle XAB=A</math>, which is an acute angle, <math>\tan{A}=t</math>, then <math>X=(1-a,at)</math>.
    5 KB (902 words) - 09:58, 20 August 2021
  • Let angle <math>\angle XAB=A</math>, which is an acute angle, <math>\tan{A}=t</math>, then <math>X=(1-a,at)</math>.
    4 KB (760 words) - 16:45, 29 April 2020
  • From Alice's point of view, <math>\tan(\theta)=\frac{z}{y}</math>. <math>\tan{30}=\frac{\sin{30}}{\cos{30}}=\frac{1}{\sqrt{3}}</math>. So, <math>y=z*\sqr From Bob's point of view, <math>\tan(\theta)=\frac{z}{x}</math>. <math>\tan{60}=\frac{\sin{60}}{\cos{60}}=\sqrt{3}</math>. So, <math>x = \frac{z}{\sqrt
    6 KB (803 words) - 00:37, 2 November 2023
  • <cmath>\frac{31}{40} \geq \tan\theta</cmath> However, <math>\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 <math>KQ=y</math>, assuming WLOG <math>x>y</math>, we must have <math>\tan(120^{\circ})=-\sqrt{3}=\dfrac{\dfrac{x+y}{100 \sqrt{3}}}{\dfrac{30000-xy}{3 ...<math>\sqrt{3}\tan{\left(\alpha\right)}</math>, we can set <math>\sqrt{3}\tan{\left(\alpha\right)}=a</math> for convenience since we really only care abo
    15 KB (2,560 words) - 01:44, 1 July 2023
  • If <math>\tan{\alpha}</math> and <math>\tan{\beta}</math> are the roots of <math>x^2 - px + q = 0</math>, and <math>\co ...cot\theta=\frac{1}{\tan\theta}</math>, we have <math>\frac{1}{\tan(\alpha)\tan(\beta)}=\frac{1}{q}=s</math>.
    1 KB (222 words) - 00:58, 20 February 2019
  • ...= \frac {VI_A}{VO} = \frac {R \sin \psi + 2R \sin \alpha}{R \cos \psi} = \tan \psi + \frac{2 \sin\alpha}{\cos \psi}.</cmath> ...cot \angle UTW = \frac {TW}{UW} = \frac {AW \cdot \tan \psi}{AU – AW} = \tan \psi \cdot \frac {2a +b+c}{b+c} =</cmath>
    6 KB (998 words) - 21:36, 17 October 2022
  • ...s clear that <math>I = \left(\frac{b + c – a}{2} , \frac{b + c – a}{2}\tan(A / 2)\right)</math>. ...and the <math>y</math> coordinate of <math>O</math> is <math>-\frac{b}{2} \tan{B-90}</math>. From this, <math>(5)</math> follows in this case as well.
    8 KB (1,449 words) - 00:09, 12 October 2023
  • ...angle at <math>C</math>. If <math>\sin A = \frac{2}{3}</math>, then <math>\tan B</math> is so <math>\tan{B} = \frac{x \sqrt{5}}{2x} = \frac{\sqrt{5}}{2}</math>, which is choice <ma
    1 KB (171 words) - 00:42, 20 February 2019
  • ...{2}+\frac{1}{3}}{1-(\frac{1}{2})(\frac{1}{3})}</math>. Simplifying, <math>\tan(\theta_a + \theta_b) = 1</math>, so <math>\theta_a + \theta_b</math> in rad
    2 KB (363 words) - 12:46, 10 May 2022
  • Let <math>f(x) = \sin{x} + 2\cos{x} + 3\tan{x}</math>, using radian measure for the variable <math>x</math>. 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 <math>f(x) = \sin{x} + 2\cos{x} + 3\tan{x}</math>, using radian measure for the variable <math>x</math>. In what in
    15 KB (2,418 words) - 16:58, 7 November 2022
  • <cmath>x(\frac {\sin \beta}{\tan{\alpha}} - \cos \beta) +x (\frac {\cos\beta}{2} +\frac{\sin\beta \sqrt{3}}{
    22 KB (3,622 words) - 17:11, 6 January 2024
  • | 67 || Senpai-Tan || 80 || 8151.479 || 101.893
    187 KB (10,824 words) - 18:27, 3 February 2022
  • If <math>\tan x=\dfrac{2ab}{a^2-b^2}</math> where <math>a>b>0</math> and <math>0^\circ <x We start by letting <math>\tan x = \frac{\sin x}{\cos x}</math> so that our equation is now: <cmath>\frac{
    1 KB (177 words) - 19:14, 2 January 2024
  • 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 <math>\tan{(2x)} = \cos{(\tfrac{x}{2})}</math> have on the interval <math>[0, 2\pi]?</
    14 KB (2,073 words) - 15:15, 21 October 2021
  • ...an use the famous mnemonic SOH CAH TOA. <math>AD=AB-DB=13-5=8 \Rightarrow \tan \angle BAC = \frac{5}{12}=\frac{r}{8} \Rightarrow 12r=40 \Rightarrow r= \fr
    5 KB (762 words) - 21:20, 20 January 2024
  • If <math>\tan a</math> and <math>\tan b</math> are the roots of <math>x^2+px+q=0</math>, 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

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