...ac {3\pi}{16} \cdot \tan^2 \frac {7\pi}{16}+\tan^2 \frac {5\pi}{16} \cdot \tan^2 \frac {7\pi}{16}?</cmath>
...ac {3\pi}{16} \cdot \tan^2 \frac {7\pi}{16}+\tan^2 \frac {5\pi}{16} \cdot \tan^2 \frac {7\pi}{16}</cmath>
...
17 KB (2,502 words) - 23:17, 7 November 2025
...a_{2023} \tan^{2023} x}{1 + a_2 \tan^2 x + a_4 \tan^4 x \cdots + a_{2022} \tan^{2022} x}
</cmath>whenever <math>\tan 2023x</math> is defined. What is <math>a_{2023}?</math>
...
9 KB (1,548 words) - 23:55, 13 October 2025
<math>\tan = \frac{\sin}{\cos}</math>
pair D = (1,tan(d));
...
19 KB (3,268 words) - 20:55, 4 February 2025
...c {\vec BD}{\vec DA} = n = \frac {\tan \alpha – \tan \gamma}{\tan \beta – \tan \gamma} > 0.</math>
...\frac {\vec AE}{\vec EC} = \frac {\tan \beta – \tan \gamma}{\tan \beta – \tan \alpha} > 0.</math>
...
59 KB (10,203 words) - 03:47, 30 August 2023
<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:
...
6 KB (1,009 words) - 19:17, 10 November 2024
...ma}, \tan \theta_C = \frac{3 – \tan \beta \cdot \tan \alpha}{\tan \beta – \tan \alpha}.</math>
...angle O'KF = \frac{3 – \tan \theta_B \cdot \tan \theta_C}{\tan \theta_C – \tan \theta_B}.</math>
...
18 KB (3,046 words) - 05:44, 19 January 2023
...\frac{s}{2} = \frac{s}{4} \cdot (sec^2(54^{\circ})-2) = \frac{s}{4} \cdot (tan^2(54^{\circ})-1)</math>.
...t in a right triangle. Thus, <math>FU = \frac{s}{4} \cdot (tan(54^{\circ})-tan(36^{\circ}))</math>
...
18 KB (2,735 words) - 23:00, 30 October 2025
...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
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17 KB (2,861 words) - 15:16, 9 August 2025
...= 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>
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5 KB (904 words) - 21:25, 19 March 2024
...ce <math>\{\theta_1, \theta_2, \theta_3...\}</math> such that <math>a_n = \tan{\theta_n}</math>, and <math>0 \leq \theta_n < 180</math>.
...+ 2}} & = \frac {\tan{\theta_n} + \tan{\theta_{n + 1}}}{1 - \tan{\theta_n}\tan{\theta_{n + 1}}} \\
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7 KB (994 words) - 15:57, 9 July 2024
...tan x+\tan y=25</math> and <math>\cot x + \cot y=30</math>, what is <math>\tan(x+y)</math>?
Since <math>\cot</math> is the reciprocal function of <math>\tan</math>:
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3 KB (545 words) - 22:44, 12 October 2023
...theta_{3}</imath>. Note that <imath>\tan(\theta_{1})=1</imath> and <imath>\tan(\theta_{3})=3</imath>, and this is why we named them as such. Let the angle
\tan(\theta_k-\theta_1)&=\tan(\theta_3-\theta_k)\\
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14 KB (2,225 words) - 19:33, 7 November 2025
...\beta)^2-\tan \alpha \tan \beta}{\tan^2 \alpha + 2\tan \alpha \tan \beta +\tan^2 \beta}</math>
...sqrt{995}</math>. We see that <math>\tan \beta = \infty</math>, and <math>\tan \alpha = \sqrt{994}</math>.
...
9 KB (1,546 words) - 13:39, 5 June 2025
...} \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
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2 KB (376 words) - 22:29, 18 May 2015
...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}}))\\
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11 KB (1,991 words) - 00:31, 19 November 2023
Suppose that <math>\sec x+\tan x=\frac{22}7</math> and that <math>\csc x+\cot x=\frac mn,</math> where <ma
...s#Pythagorean Identities|trigonometric Pythagorean identities]] <math>1 + \tan^2 x = \sec^2 x</math> and <math>1 + \cot^2 x = \csc^2 x</math>.
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10 KB (1,686 words) - 23:29, 28 June 2025
...leq y \leq 100 \pi</math>, <math>x + y = 20.19</math>, and <math>\tan x + \tan y = 20.19</math>?
According to the <imath>\tan</imath> angle sum trigonometric identity,
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2 KB (280 words) - 08:58, 7 November 2025
<cmath>\tan \angle APX = \tan \angle SPY \implies \angle APX = \angle SPY \implies XY || AP.</cmath>
<cmath>\angle BI_CC' = \beta, B'C = BC' = BD = s - a = R \tan \beta,</cmath>
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122 KB (21,381 words) - 13:47, 3 August 2025
* <math>\tan^2x + 1 = \sec^2x</math>
* <math>\tan(x + y) = \frac{\tan (x) + \tan (y)}{1 - \tan (x) \tan (y)} </math>
...
8 KB (1,397 words) - 20:55, 20 January 2024
...ective medians; in other words, <math>\tan \theta_1 = 1</math>, and <math>\tan \theta_2 =2</math>.
...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 (1,755 words) - 17:58, 3 June 2025
