Question

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

Answer 1

`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`