Note that
[tex] \frac{17 \pi }{12} = \frac{3 \pi }{12} + \frac{14 \pi }{12} = \frac{ \pi }{4} + \frac{7 \pi }{6} [/tex]
Note that
[tex]x= \frac{ \pi }{4}:\,\, sin(x) =cos(x)= \frac{1}{ \sqrt{2} },\,tan(x)=1\\x= \frac{7 \pi }{6} :\,\,sin(x)=- \frac{1}{2} ,\,\,cos(x)=- \frac{ \sqrt{3} }{2} ,\,\,tan(x)= \frac{1}{ \sqrt{3} } [/tex]
Use the identity
[tex]tan(x+y)= \frac{tan(x)+tan(y)}{1-tan(x)tan(y)} [/tex]
Therefore
[tex]tan( \frac{17 \pi }{12} )= \frac{1+ \frac{1}{ \sqrt{3} } }{1- \frac{1}{ \sqrt{3} } } = \frac{ \sqrt{3}+1 }{ \sqrt{3}-1} = \frac{( \sqrt{3}+1 )^{2}}{( \sqrt{3}-1 )( \sqrt{3}+1 )} = \frac{3+1+2 \sqrt{3}}{3-1} =2+ \sqrt{3}[/tex]
Answer: 2 + √3
