Question

What is the tangent of -pi/12. please explain

Answer 1

`2\cdot \cfrac{\pi }{12}\implies \cfrac{\pi }{6}\hspace{5em}therefore\hspace{5em}\cfrac{~~ ( \pi )/( 6 ) ~~}{2}\implies \cfrac{\pi }{12} \\\\[-0.35em] ~\dotfill\\\\ \tan\left(\cfrac{\theta}{2}\right)= \begin{cases} \pm \sqrt{\cfrac{1-\cos(\theta)}{1+\cos(\theta)}} \\\\ \cfrac{\sin(\theta)}{1+\cos(\theta)} \\\\ \cfrac{1-\cos(\theta)}{\sin(\theta)}\leftarrow \textit{we'll use this one} \end{cases} \\\\[-0.35em] ~\dotfill`

`\tan\left( \cfrac{\pi }{12} \right)\implies \tan\left( \cfrac{~~ ( \pi )/( 6 ) ~~}{2} \right)=\cfrac{1-\cos\left( (\pi )/(6) \right)}{\sin\left( (\pi )/(6) \right)}`

`\tan\left( \cfrac{~~ ( \pi )/( 6 ) ~~}{2} \right)=\cfrac{ ~~ 1-(√(3))/(2) ~~ }{(1)/(2)}\implies \tan\left( \cfrac{~~ ( \pi )/( 6 ) ~~}{2} \right)=\cfrac{~~ ( 2-√(3) )/( 2 ) ~~}{(1)/(2)} \\\\\\ \stackrel{ \textit{this is for the 1st Quadrant} }{\tan\left( \cfrac{\pi }{12} \right)=2-√(3)}\hspace{5em} \stackrel{ \textit{on the IV Quadrant, tangent is negative} }{\tan\left( -\cfrac{\pi }{12} \right)=√(3)-2}`