How do you evaluate tan (11Pi/12)?

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How do you evaluate tan (11Pi/12)?

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Answer 1 · 2016-07-03T11:35:21

$tan(11pi/12)=-tan(pi/12)=sqrt3-2~=1.7321-2=-0.2679.$

Explanation:

$tan(11pi/12)=tan(pi-pi/12)=-tan(pi/12)=-tantheta$ , say where, $theta=pi/12$ , so that $2theta=pi/6.$

Now recall the identity $: tan2theta=(2tantheta)/(1-tan^2theta)$

$:. tan(pi/6)=(2t)/(1-t^2)$ , where, $t=tantheta=tan(pi/12)$
$:. 1/sqrt3=(2t)/(1-t^2) rArr1-t^2=2sqrt3*trArrt^2+2sqrt3*t=1rArrt^2+2sqrt3*t+(sqrt3)^2=1+(sqrt3)^2$
$:. (t+sqrt3)^2=2^2$
$:.t+sqrt3=+-2$
$:.t=+-2-sqrt3$

$t=-2-sqrt3 rArr tan(pi/12)$ is $-ve$ , which is not possible, as $pi/12$ lies in the First Quadrant [all trigo. ratio $+ve$ ]

Hence, $t=tan(pi/12)=+2-sqrt3$ , so, $tan(11pi/12)=-tan(pi/12)=sqrt3-2~=1.7321-2=-0.2679$

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How do you evaluate tan (11Pi/12)?

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